Free math worksheets

- Start from the general exponential form \(y=ab^x\). - Apply a logarithm to turn multiplication and exponents into addition and multiplication. - Identify which original variable remains unchanged in the resulting linear equation.

1. An exponential model can be written as \(y=ab^x\), where \(y>0\). 2. Take the natural logarithm of the response: \(\ln(y)=\ln(a)+x\ln(b)\). 3. This equation is linear in \(x\). Use transformed explanatory variable \(x^*=x\) and transformed response variable \(y^*=\ln(y)\).

Use \(x^*=x\) and \(y^*=\ln(y)\); plot \(\ln(y)\) against \(x\). Then \(y=ab^x\) becomes \(\ln(y)=\ln(a)+x\ln(b)\).

- Begin with the power form \(y=ax^b\). - Log both sides so the exponent becomes a coefficient. - Identify the transformed variable represented on each axis of the linear plot.

1. A power model has the form \(y=ax^b\). 2. Take the natural logarithm of both variables: \(\ln(y)=\ln(a)+b\ln(x)\). 3. This equation is linear in \(\ln(x)\). Use transformed explanatory variable \(x^*=\ln(x)\) and transformed response variable \(y^*=\ln(y)\).

Use \(x^*=\ln(x)\) and \(y^*=\ln(y)\); plot \(\ln(y)\) against \(\ln(x)\). Then \(y=ax^b\) becomes \(\ln(y)=\ln(a)+b\ln(x)\).

- Write the general logarithmic model \(y=a+b\ln(x)\). - Identify which variable appears inside the logarithm. - The response does not need to be transformed for this model form.

1. A logarithmic model has the form \(y=a+b\ln(x)\). 2. The response \(y\) is already linear in \(\ln(x)\). 3. Use transformed explanatory variable \(x^*=\ln(x)\) and transformed response variable \(y^*=y\). Plot \(y\) against \(\ln(x)\).

Use \(x^*=\ln(x)\) and \(y^*=y\); plot \(y\) against \(\ln(x)\). The model \(y=a+b\ln(x)\) is linear in the transformed explanatory variable.

- Identify whether the response variable was transformed. - A regression of \(y\) on \(\ln(x)\) already predicts \(y\) on its original scale. - Match the form \(a+b\ln(x)\) to its model type.

1. Only the explanatory variable was transformed; the response \(y\) remained on its original scale. 2. Therefore, no back-transformation of \(\hat y\) is needed. 3. The original-variable model is \(\hat y=7.00-0.70\ln(x)\), which is a logarithmic model.

\(\hat y=7.00-0.70\ln(x)\), a logarithmic model.

- First calculate the predicted value of \(\ln(y)\) at \(x=2\). - Exponentiate to undo the natural logarithm. - Round only after returning to the original response scale.

1. Substitute \(x=2\) into the transformed model: \(\ln(\hat y)=0.500-0.180(2)=0.140\). 2. Exponentiate to return to the original response scale: \(\hat y=e^{0.140}\approx1.150\).

\(\hat y\approx1.150\).

- Substitute \(4\) for \(x\) directly in the logarithmic model. - The response is already on its original scale, so do not exponentiate. - Round after evaluating the full expression.

1. Substitute \(x=4\) into the model: \(\hat y=14.00-1.40\ln(4)\). 2. Evaluate: \(\hat y\approx14.00-1.40(1.386294)\approx12.059\). 3. The predicted resale price is approximately \(12.059\) thousand dollars.

\(\hat y\approx12.059\), or approximately \(\$12{,}059\).

- Compare the magnitudes of the three correlation coefficients. - Identify which pair of transformed variables corresponds to the largest \(|r|\). - A linear model for \(\ln(y)\) versus \(x\) becomes an exponential model for \(y\).

1. Compare the absolute correlation values: \(|0.61|=0.61\), \(|0.97|=0.97\), and \(|0.72|=0.72\). 2. The value \(0.97\) is closest to \(1\), so the plot of \(\ln(y)\) against \(x\) is most nearly linear. 3. A linear relationship between \(\ln(y)\) and \(x\) suggests an exponential model for \(y\) as a function of \(x\).

The \((x,\ln(y))\) relationship is most nearly linear because \(|r|=0.97\) is largest. It suggests an exponential model on the original scale.

- Compare the magnitudes, not the signs, of the correlations. - Identify the transformation pair with \(|r|\) closest to \(1\). - A linear log-log relationship corresponds to a power model.

1. Compare the absolute correlation values: \(|0.88|=0.88\), \(|0.75|=0.75\), and \(|0.99|=0.99\). 2. The value \(0.99\) is closest to \(1\), so the plot of \(\ln(y)\) against \(\ln(x)\) is most nearly linear. 3. A linear relationship between \(\ln(y)\) and \(\ln(x)\) suggests a power model for \(y\) as a function of \(x\).

The \((\ln(x),\ln(y))\) relationship is most nearly linear because \(|r|=0.99\) is largest. It suggests a power model on the original scale.

- Compare the magnitudes of the negative correlations. - Identify which transformed relationship has \(|r|\) closest to \(1\). - A negative linear relationship between \(x\) and \(\ln(y)\) corresponds to exponential decay.

1. Compare the absolute correlation values: \(|-0.83|=0.83\), \(|-0.98|=0.98\), and \(|-0.90|=0.90\). 2. The value \(0.98\) is closest to \(1\), so the plot of \(\ln(y)\) against \(x\) is most nearly linear. 3. A negative linear relationship between \(\ln(y)\) and \(x\) suggests an exponential decay model for \(y\) as a function of \(x\).

The \((x,\ln(y))\) relationship is most nearly linear because \(|r|=0.98\) is largest. The negative correlation suggests an exponential decay model.

- Match each transformation to a specific model form. - Compare exponential linearization with power-model linearization. - A transformation is justified by the resulting plot and residuals, not merely by the fact that the original data are curved.

1. A response-log transformation linearizes model forms in which \(\ln(y)\) is linear in \(x\), such as exponential models. 2. It does not linearize every curved relationship. For example, a power relationship requires a log-log plot of \(\ln(y)\) against \(\ln(x)\). 3. The transformed scatterplot and residual plot must be checked to determine whether the chosen transformation actually produces a suitable linear model.

Taking \(\ln(y)\) linearizes only certain model forms, such as exponential relationships. Other curves may require a different transformation, and the transformed scatterplot and residuals must still be checked.

- Exponentiation must undo the logarithm applied to the entire expression. - Use \(e^{u+v}=e^u e^v\), not \(e^{u+v}=u+e^v\). - Check the model at \(x=0\): \(\ln(y)=2\) requires \(y=e^2\), not \(3\).

1. Exponentiate the entire right side: \(y=e^{2+0.4x}\). 2. Use the exponent rule \(e^{u+v}=e^u e^v\): \(y=e^2e^{0.4x}=e^2(e^{0.4})^x\). 3. Therefore, the original model is exponential, \(y\approx7.389(1.492)^x\), not a sum of \(2\) and an exponential term.

The correct model is \(y=e^{2+0.4x}=e^2(e^{0.4})^x\approx7.389(1.492)^x\), not \(y=2+e^{0.4x}\).

- Match \(\ln(y)=A+B\ln(x)\) with \(\ln(a)+b\ln(x)\). - The transformed intercept is \(\ln(a)\). - Exponentiate the intercept while keeping the transformed slope as the power exponent.

1. The transformed intercept \(1.1\) equals \(\ln(a)\), not \(a\). 2. Exponentiate the intercept: \(a=e^{1.1}\approx3.004\). 3. The transformed slope is the power exponent, \(b=1.6\). Therefore, the original-scale model is \(y=e^{1.1}x^{1.6}\approx3.004x^{1.6}\).

The coefficient is \(a=e^{1.1}\approx3.004\), not \(1.1\). The original-scale model is \(y\approx3.004x^{1.6}\).

- Compare the strength of \(r=0.95\) with \(r=-0.98\). - Correlation measures linear association on the transformed scale but does not check residual patterns. - Use the known circle-area relationship to identify the scientifically appropriate model form.

1. The strength of a linear relationship depends on \(|r|\), not on whether \(r\) is positive. A correlation near \(-1\) is just as strongly linear as one near \(1\). 2. Even the largest \(|r|\) does not by itself guarantee that a transformed model is appropriate. The residual plot, domain restrictions, predictive performance, and mathematical or scientific context must also support the model. 3. For radius and circle area, the known relationship \(y=\pi x^2\) is a power model, so a log-log transformation is justified by both the model form and the data diagnostics.

The statement is incorrect. Compare correlations by \(|r|\), because strong negative and positive correlations can both indicate linearity. Then use residual plots and context to determine whether the transformation is appropriate; for circle area and radius, the power relationship \(y=\pi x^2\) supports a log-log model.

- Start from the multiplicative form of an exponential model. - Think about what taking a logarithm does to an exponent. - Relate a multiplicative factor below \(1\) to the direction of change.

1. For an exponential model \(\hat y=ab^x\), taking natural logs gives \(\ln(\hat y)=\ln(a)+x\ln(b)\). 2. The transformed slope is therefore \(\ln(0.88)\approx-0.128\). 3. The slope is negative because a factor less than \(1\) represents exponential decay as \(x\) increases.

The transformed slope is approximately \(-0.128\). It is negative because the response is multiplied by a factor less than \(1\) for each increase in \(x\).

- First make the prediction on the scale used by the fitted linear model. - Then undo the transformation to return to the original response scale. - Check that the back-transformed prediction is positive.

1. At \(x=20\), the transformed fitted value is \(\hat v=0.020+0.005\cdot20=0.120\). 2. Because \(v=\frac{1}{y}\), return to the original scale with \(\hat y=\frac{1}{\hat v}\). 3. Thus \(\hat y=\frac{1}{0.120}\approx8.33\).

The predicted original response is approximately \(8.33\).

- Replace the transformed variable by its definition in the fitted equation. - Compare what happens to the transformed value when the sign of \(x\) changes. - Use the rewritten model for both requested predictions.

1. Substitute \(u=x^2\) into the transformed linear model to get \(\hat y=7+0.4x^2\). 2. At \(x=-3\), \(\hat y=7+0.4\cdot9=10.6\). 3. At \(x=3\), the squared explanatory value is also \(9\), so \(\hat y=10.6\). 4. The transformation produces an original-scale model symmetric about \(x=0\).

The original-scale model is \(\hat y=7+0.4x^2\). It predicts \(10.6\) at both \(x=-3\) and \(x=3\).

- Replace the transformed predictor by its definition. - Then consider what happens to the transformed term when the original explanatory value becomes very large. - The constant term determines the long-run baseline once the transformed contribution becomes negligible.

1. Substitute \(u=\frac{1}{x^2}\) to obtain \(\hat y=5+\frac{120}{x^2}\). 2. As \(x\) becomes very large, \(\frac{120}{x^2}\) approaches \(0\). 3. Therefore, the predicted response approaches \(5\).

The original-scale model is \(\hat y=5+\frac{120}{x^2}\), and its predicted response approaches \(5\) as \(x\) increases.

- Undo the natural logarithm by exponentiating both sides. - Use \(e^{u+v}=e^u e^v\). - Match \(e^{0.800}(e^{0.120})^x\) with the form \(ab^x\).

1. Exponentiate both sides: \(\hat y=e^{0.800+0.120x}\). 2. Separate the factors: \(\hat y=e^{0.800}(e^{0.120})^x\). 3. Compute the constants: \(a=e^{0.800}\approx2.226\), \(b=e^{0.120}\approx1.127\). 4. Therefore, \(\hat y\approx2.226(1.127)^x\).

\(\hat y\approx2.226(1.127)^x\).

- Exponentiate to undo the natural logarithm. - Separate the exponential of a sum into a product. - Use the identity \(e^{B\ln(x)}=x^B\).

1. Exponentiate both sides: \(\hat y=e^{1.250+1.320\ln(x)}\). 2. Separate the factors: \(\hat y=e^{1.250}e^{1.320\ln(x)}\). 3. Use \(e^{B\ln(x)}=x^B\): \(\hat y=e^{1.250}x^{1.320}\). 4. Since \(e^{1.250}\approx3.490\), the model is \(\hat y\approx3.490x^{1.320}\).

\(\hat y\approx3.490x^{1.320}\).

- Substitute \(x=3\) into the log-log regression equation. - Keep extra decimal places in the transformed prediction. - Exponentiate once to return to the original response scale, then round.

1. Substitute \(x=3\): \(\ln(\hat y)=0.820+0.900\ln(3)\approx1.808751\). 2. Exponentiate to return to the original response scale: \(\hat y=e^{1.808751}\approx6.103\).

\(\hat y\approx6.103\).

- Substitute \(x=6\) into the log-log regression equation. - Keep extra decimal places in the transformed prediction. - Exponentiate once, then round on the original scale.

1. Substitute \(x=6\): \(\ln(\hat y)=1.180+1.200\ln(6)\approx3.330111\). 2. Exponentiate to return to the original response scale: \(\hat y=e^{3.330111}\approx27.941\). 3. The predicted number of visits is approximately \(27.941\) thousand.

\(\hat y\approx27.941\), or approximately \(27{,}941\) visits.

- Match \(\ln(\hat y)=A+Bx\) with \(\hat y=e^A(e^B)^x\). - The initial value is the prediction at \(x=0\). - Convert the multiplicative factor \(b\) to a percent change using \((b-1)100\%\).

1. The original model has the form \(\hat y=ab^x\), where \(a=e^{1.200}\) and \(b=e^{0.095}\). 2. Compute the initial value: \(a=e^{1.200}\approx3.320\). 3. Compute the one-unit multiplicative factor: \(b=e^{0.095}\approx1.100\). 4. The percent change per unit is \((b-1)100\%\approx9.9659\%\approx10.0\%\). The model predicts that the bacteria count is multiplied by about \(1.100\), an increase of about \(10.0\%\), for each one-unit increase in time.

\(a\approx3.320\); multiplicative factor \(\approx1.100\). The predicted bacteria count increases by about \(10.0\%\) per one-unit increase in time.

- Match \(\ln(\hat y)=A+Bx\) with \(\hat y=e^A(e^B)^x\). - A negative transformed slope produces a factor between \(0\) and \(1\). - Convert a decay factor \(b\) to a percent decrease using \((1-b)100\%\).

1. The original model has the form \(\hat y=ab^x\), where \(a=e^{0.400}\) and \(b=e^{-0.220}\). 2. Compute the initial value: \(a=e^{0.400}\approx1.492\). 3. Compute the one-unit multiplicative factor: \(b=e^{-0.220}\approx0.803\). 4. The percent decrease per unit is \((1-b)100\%\approx19.7481\%\approx19.7\%\). The model predicts that the medication concentration is multiplied by about \(0.803\), a decrease of about \(19.7\%\), for each one-unit increase in time.

\(a\approx1.492\); multiplicative factor \(\approx0.803\). The predicted medication concentration decreases by about \(19.7\%\) per one-unit increase in time.

- Exponentiate the intercept to obtain the original-scale coefficient. - The slope in a log-log model becomes the power exponent. - In a power model, interpret the exponent as the approximate percent change in \(y\) for a \(1\%\) change in \(x\).

1. Exponentiate the transformed equation: \(\hat y=e^{0.80}x^{1.25}\). 2. The coefficient is \(a=e^{0.80}\approx2.226\), and the exponent is \(b=1.25\). 3. In a power model, the exponent is the elasticity. A \(1\%\) increase in advertising spending is associated with approximately a \(1.25\%\) increase in predicted sales.

\(\hat y\approx2.226x^{1.25}\). The exponent \(1.25\) means that a \(1\%\) increase in advertising spending is associated with approximately a \(1.25\%\) increase in predicted sales.

- Check for an exact tie before naming one transformation. - Map \((\ln(x),y)\) to a logarithmic model and \((\ln(x),\ln(y))\) to a power model. - When correlations are tied, compare residual behavior and scientific plausibility.

1. Compare the absolute correlations. The \((\ln(x),y)\) and \((\ln(x),\ln(y))\) relationships tie at \(|r|=0.95\), while \((x,\ln(y))\) has \(|r|=0.94\). 2. The tie means correlation alone does not identify a unique best transformation. 3. The tied candidates suggest a logarithmic model and a power model, respectively. Residual plots, model assumptions, predictive performance, and scientific context are needed to choose between them.

There is no unique choice from the correlations alone. The \((\ln(x),y)\) and \((\ln(x),\ln(y))\) relationships tie at \(|r|=0.95\), suggesting logarithmic and power models, respectively. Use residual plots and context to decide.

- Think about how to reverse the transformation applied to the response variable. - Separate the constant part from the part that depends on the explanatory variable. - In an exponential model, interpret the base as a multiplicative change rather than an additive change.

1. Undo the common logarithm: \(\hat y=10^{1.200+0.080x}=10^{1.200}(10^{0.080})^x\). 2. The coefficient is \(a=10^{1.200}\approx15.849\), and the growth factor is \(b=10^{0.080}\approx1.202\). 3. Each one-unit increase in \(x\) multiplies the predicted response by about \(1.202\), corresponding to an increase of about \(20.2\%\).

\(\hat y\approx15.849(1.202)^x\). For each one-unit increase in \(x\), the predicted response is multiplied by about \(1.202\), or increases by about \(20.2\%\).

- Relate common logarithms and natural logarithms through a change of base. - Apply the conversion to the entire transformed equation. - Check that both coefficients change consistently.

1. Since \(\ln(y)=\ln(10)\log_{10}(y)\), multiply the entire common-log equation by \(\ln(10)\). 2. This gives \(A=1.300\ln(10)\approx2.993\). 3. The new slope is \(B=0.200\ln(10)\approx0.461\).

\(\ln(\hat y)\approx2.993+0.461x\).

- Put the requested original-scale response into the transformed equation first. - Isolate the transformed explanatory variable. - Reverse the explanatory-variable transformation only after solving on the transformed scale.

1. Set the transformed predicted response equal to \(\ln(20)\): \(\ln(20)=1.100+1.400\ln(x)\). 2. Solve for the transformed explanatory value: \(\ln(x)=\frac{\ln(20)-1.100}{1.400}\). 3. Back-transform to obtain \(x=\exp\left(\frac{\ln(20)-1.100}{1.400}\right)\approx3.873\).

\(x\approx3.873\).

- Write the transformed residual as the difference between the transformed observed and predicted responses. - Use a logarithm property to combine the difference into a ratio. - Reverse the logarithm before interpreting the result on the original response scale.

1. A transformed residual of \(0.200\) means \(\ln(y)-\ln(\hat y)=0.200\). 2. Therefore, \(\ln\left(\frac{y}{\hat y}\right)=0.200\), so \(\frac{y}{\hat y}=e^{0.200}\approx1.221\). 3. Thus, the observed response is about \((1.221-1)\cdot100\%=22.1\%\) larger than the predicted response.

The observed response is about \(1.221\) times the predicted response, or about \(22.1\%\) larger.

- Check the domain of every transformation before applying it to the data. - Look specifically at the smallest explanatory value in the data set. - Ask whether changing the variable before taking a logarithm leaves the original model family unchanged.

1. The proposed log-log transformation requires positive explanatory values because \(\ln(x)\) is defined only for \(x>0\) in this context. 2. The observation with \(x=0\) cannot be transformed using \(\ln(x)\), so the proposed transformation cannot be applied to all five observations as stated. 3. Replacing \(x\) with \(x+1\) creates a different transformed variable and therefore a different original-scale model. It is not a harmless way to fit the same power model. 4. Any shifted transformation would need a substantive modeling justification and a new interpretation.

No. The observation with \(x=0\) makes \(\ln(x)\) undefined, so the log-log transformation cannot be applied to all five observations as stated. Replacing \(x\) by \(x+1\) changes the model rather than preserving the original power-model interpretation.

- Recall how logarithms with different bases are related. - Ask whether multiplying all transformed response values by one positive constant changes the shape of their association with \(x\). - Separate a change in vertical scale from a change in model form.

1. The two transformed responses are related by \(\ln(y)=\ln(10)\log_{10}(y)\). 2. Thus, changing from common logarithms to natural logarithms multiplies every transformed response by the same positive constant. 3. A positive linear rescaling changes the vertical scale and fitted coefficients but does not change whether the transformed association is linear or the pattern of residuals apart from scale. 4. Therefore, the logarithm base should not be chosen by looking for a straighter pattern; either base represents the same linearizing transformation.

The statement is incorrect. Natural logs and common logs differ only by a positive constant factor, so they produce the same linearity pattern. The axis scale and regression coefficients change, but one base cannot make the relationship intrinsically straighter than the other.

- Express the new response as a constant multiple of the old response. - Use a logarithm property to separate the constant factor from the original transformed response. - Decide whether the conversion changes the part of the model attached to \(x\).

1. Since \(y_c=100y\), \(\ln(y_c)=\ln(100)+\ln(y)\). 2. Therefore, \(\ln(\hat y_c)=\ln(100)+0.800+0.120x\approx5.405+0.120x\). 3. The transformed slope remains \(0.120\) because multiplying the original response by a positive constant only shifts its logarithm by a constant. 4. The original-scale multiplicative factor per one-unit increase in \(x\) is still \(e^{0.120}\), so the growth rate is unchanged.

\(\ln(\hat y_c)\approx5.405+0.120x\). The transformed slope remains \(0.120\), and the original-scale multiplicative factor per one-unit increase in \(x\) is unchanged.

- Do not choose a transformation from one summary statistic alone. - Compare the residual patterns after each transformation. - Identify the original-scale model family associated with transforming both positive variables logarithmically.

1. A large transformed correlation alone is not enough to establish that a linear model is appropriate. 2. Model A has residual curvature, so its transformed relationship still has systematic nonlinearity. 3. Model B has a slightly smaller transformed correlation but a residual plot consistent with linear form and roughly constant spread. 4. A linear relationship between \(\ln(y)\) and \(\ln(x)\) corresponds to a power model on the original scale.

Model B is better supported. Its residual plot is consistent with a linear model, and the log-log transformation suggests an original-scale power model.

- Compare the model at \(x\) and at a constant multiple of \(x\). - Use a logarithm property to separate the multiplier from the original explanatory value. - Subtract the two predictions and look for terms that cancel.

1. At explanatory value \(x\), the prediction is \(12.0+4.0\ln(x)\). 2. At \(3x\), the prediction is \(12.0+4.0\ln(3x)=12.0+4.0\ln(3)+4.0\ln(x)\). 3. The predicted change is therefore \(4.0\ln(3)\approx4.394\), independent of the starting value of \(x\).

Multiplying \(x\) by \(3\) increases the predicted response by approximately \(4.394\) units.

- Compare the ratio of the two predictions with the number of explanatory-variable steps between them. - In an exponential model, equal steps in \(x\) multiply predictions by the same factor. - Connect the original-scale base to the slope after taking natural logarithms.

1. From \(x=2\) to \(x=5\), the explanatory variable increases by \(3\), so the predicted response is multiplied by \(b^3\). 2. Thus, \(b^3=\frac{54.0}{20.0}=2.7\), giving \(b=2.7^{1/3}\approx1.392\). 3. In the log-linear form, the slope is \(B=\ln(b)\approx\ln(1.39247665)\approx0.331\).

The one-unit multiplicative factor is \(b\approx1.392\), and the slope of the linearized model is \(B\approx0.331\).

- Express the original explanatory variable in terms of the new unit. - Substitute before simplifying the power. - Distinguish the coefficient of a power model from its exponent.

1. Since \(u=100x\), \(x=\frac{u}{100}\). 2. Substituting gives \(\hat y=3.2\left(\frac{u}{100}\right)^{1.5}=\frac{3.2}{100^{1.5}}u^{1.5}=0.0032u^{1.5}\). 3. The exponent remains \(1.5\), so the slope of a regression of \(\ln(y)\) on \(\ln(u)\) remains \(1.5\). 4. Changing the explanatory-variable unit changes the coefficient but not the power exponent.

\(\hat y=0.0032u^{1.5}\). The log-log slope remains \(1.5\).

- Identify which quantity is actually zero at the transformed intercept. - Solve for the original explanatory value that makes its logarithm zero. - Reverse the response logarithm to return to the original scale.

1. The transformed explanatory variable is \(\ln(x)\), so the intercept applies when \(\ln(x)=0\). 2. The equation \(\ln(x)=0\) gives \(x=1\), not \(x=0\). 3. At \(x=1\), \(\ln(\hat y)=1.500\), so \(\hat y=e^{1.500}\approx4.482\). 4. Thus, \(e^{1.500}\) is the coefficient in the original power model and the predicted response at \(x=1\).

The intercept applies at \(x=1\), because \(\ln(1)=0\). The predicted response there is \(e^{1.500}\approx4.482\).

- Compare the model at \(x\) and at \(2x\) using a ratio. - Express the stated multiplicative change as a power of \(2\). - Recall what the exponent becomes after taking logarithms of a power model.

1. In a power model, doubling \(x\) multiplies the prediction by \(2^b\). 2. The given condition therefore gives \(2^b=5\). 3. Taking logarithms gives \(b=\frac{\ln(5)}{\ln(2)}\approx2.322\). 4. In the log-log linearized model, the slope is the power exponent, so the transformed slope is also approximately \(2.322\).

\(b\approx2.322\). The slope of the log-log regression is also approximately \(2.322\).

- Express the old time variable in terms of the shifted one. - Separate the constant power of the exponential base from the part involving the new variable. - A shift in the time origin changes the starting level but not the growth factor per time step.

1. Since \(u=x-3\), \(x=u+3\). 2. Substituting gives \(\hat y=5.00(1.20)^{u+3}=5.00\cdot(1.20)^3(1.20)^u=8.64(1.20)^u\). 3. The one-year multiplicative factor remains \(1.20\) because shifting the time origin does not change the growth rate. 4. The log-linear slope remains \(\ln(1.20)\approx0.182\); only the intercept changes.

\(\hat y=8.64(1.20)^u\). The multiplicative factor remains \(1.20\), and the log-linear slope remains approximately \(0.182\).

- Replace each explanatory value by its reciprocal and compare how the response changes across the transformed pairs. - Look for a constant rate of change after the transformation. - For the prediction, transform the new explanatory value before using the linear relationship.

1. The transformed explanatory values are \(u=\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}\). 2. For each transformed point, \(y-3=24u\), so the points lie on the line \(\hat y=3+24u\). 3. When \(x=8\), \(u=\frac{1}{8}\). 4. The prediction is \(\hat y=3+24\cdot\frac{1}{8}=6\).

The reciprocal transformation gives the linear model \(\hat y=3+24u\), where \(u=\frac{1}{x}\). For \(x=8\), the predicted response is \(6\).

- Evaluate the transformed explanatory value at the three requested \(x\)-values. - Compare changes on the transformed scale before comparing predictions. - Interpret the slope with respect to the variable that appears linearly in the fitted equation.

1. The predictions are \(11\) at \(x=1\), \(14\) at \(x=4\), and \(17\) at \(x=9\). 2. The predicted response increases by \(3\) from \(1\) to \(4\) and by \(3\) from \(4\) to \(9\). 3. In both cases, \(\sqrt{x}\) increases by \(1\), even though the raw increases in \(x\) are different. 4. Thus, the coefficient \(3\) is the change in predicted response for a one-unit increase in \(\sqrt{x}\), not for a one-unit increase in \(x\).

Each interval increases the prediction by \(3\). The coefficient \(3\) applies to a one-unit increase in \(\sqrt{x}\), so the model does not imply a constant change in \(y\) for equal raw changes in \(x\).

- Identify the response variable used to compute each \(R^2\). - Ask whether the unexplained variation is being measured on the same scale. - Consider what must happen before predictions from the transformed model can be evaluated on the original scale.

1. The two \(R^2\) values describe variation on different response scales: \(y\) for the first model and \(\ln(y)\) for the second. 2. A larger \(R^2\) on the transformed scale does not directly measure smaller prediction error on the original \(y\)-scale. 3. Model comparison should also consider residual patterns on the fitted scale, the appropriateness of the transformation, and prediction performance after back-transformation.

No. Because the models use different response scales, their \(R^2\) values do not directly compare prediction accuracy for the original response \(y\).

- A log-log slope describes how multiplicative changes in \(x\) affect multiplicative changes in \(y\). - Convert the \(25\%\) increase into a factor before applying the model's exponent. - Turn the resulting response factor back into a percent change at the end.

1. A \(25\%\) increase multiplies \(x\) by \(1.25\). 2. In the corresponding power model, the predicted response is multiplied by \(1.25^{0.8}\approx1.19544\). 3. Therefore, the predicted response increases by approximately \((1.19544-1)\cdot100\%\approx19.54\%\).

The predicted response increases by approximately \(19.54\%\).

- Apply the same logarithm to the entire exponential model. - Use logarithm rules to separate the constant multiplier from the exponential term. - The exponent becomes a multiplier after taking the logarithm.

1. Take the base-10 logarithm of both sides: \(\log_{10}(\hat y)=\log_{10}(5)+x\log_{10}(1.2)\). 2. Numerically, \(\log_{10}(5)\approx0.69897\) and \(\log_{10}(1.2)\approx0.07918\). 3. Therefore, the transformed linear model is \(\log_{10}(\hat y)\approx0.69897+0.07918x\).

\(\log_{10}(\hat y)\approx0.69897+0.07918x\).

- Treat \(\ln(y)\) as the vertical coordinate when finding the transformed-line slope. - Use one transformed point to solve for \(A\). - Convert \(A\) and \(B\) with \(a=e^A\) and \(b=e^B\).

1. Find the slope of the transformed line: \(B=\frac{\ln(10.985)-\ln(5.000)}{4-1}\approx0.262364\). 2. Use the point \((1,\ln(5.000))\) to find the intercept: \(A=\ln(5.000)-0.262364(1)\approx1.347074\). 3. Back-transform the parameters: \(a=e^A\approx3.846\), \(b=e^B=1.300\). 4. Therefore, the exponential model is \(y\approx3.846(1.300)^x\).

\(y\approx3.846(1.300)^x\).

- On a log-log line, the slope equals the power exponent \(b\). - Use one transformed point to solve for \(\ln(a)\). - Exponentiate \(\ln(a)\) to recover the original-scale coefficient.

1. The slope of the log-log line is the power exponent: \(b=\frac{\ln(22.229)-\ln(6.762)}{\ln(7)-\ln(2)}\approx0.949963\). 2. Use the first point to find the transformed intercept: \(\ln(a)=\ln(6.762)-0.949963\ln(2)\approx1.252854\). 3. Back-transform the intercept: \(a=e^{1.252854}\approx3.500\). 4. Therefore, the power model is \(y\approx3.500x^{0.950}\).

\(y\approx3.500x^{0.950}\).

- Compare two predictions by forming a ratio rather than finding either prediction separately. - In the transformed model, a change in the explanatory variable becomes a multiplicative change on the original response scale. - Use the inverse of the transformation once you have an equation for the desired ratio.

1. For a change of \(\Delta x\), the ratio of the new prediction to the old prediction is \(e^{-0.180\Delta x}\). 2. Set the ratio equal to \(0.5\): \(e^{-0.180\Delta x}=0.5\). 3. Taking natural logarithms gives \(-0.180\Delta x=\ln(0.5)\), so \(\Delta x=\frac{\ln(0.5)}{-0.180}\approx3.851\).

\(x\) must increase by about \(3.851\) units.