Free math worksheets

- Rewrite the logarithmic statement as an equivalent exponential statement. - The base is \(4\), and the logarithm value is \(2\).

Rewrite the logarithmic equation in exponential form: \(x=4^2\). Therefore, \(x=16\).

\(x=16\)

- Solving the equation graphically means finding where the two displayed y-values agree. - Locate the intersection of the logarithmic curve and the horizontal line. - Read the x-coordinate to the requested precision.

1. A solution occurs where the logarithmic curve and the horizontal line intersect. 2. The intersection has x-coordinate about \(1.6\). 3. Therefore, \(\ln(x)=0.5\) has solution \(x\approx1.6\) to the nearest tenth.

\(x\approx1.6\)

- Rewrite the logarithmic statement as an equivalent exponential equation. - Express both sides using the same base. - Remember that a logarithm base may lie between \(0\) and \(1\).

1. Rewrite the logarithmic equation in exponential form: \(\left(\frac{1}{3}\right)^x=27\). 2. Since \(27=3^3\) and \(\frac{1}{3}=3^{-1}\), the equation becomes \(3^{-x}=3^3\). 3. Therefore, \(-x=3\), so \(x=-3\).

\(x=-3\)

- Set the logarithm input greater than zero. - Isolate the logarithm. - Use the exponential function to undo \(\ln\). - Check the solution against the domain.

1. The logarithm requires \(2x-4>0\), so the domain is \((2, \infty)\). 2. Isolate the logarithm: \(3\ln(2x-4)=6\), so \(\ln(2x-4)=2\). 3. Exponentiate: \(2x-4=e^2\). 4. Therefore, \(x=2+\frac{e^2}{2}\), which lies in the domain.

Domain: \((2, \infty)\); \(x=2+\frac{e^2}{2}\)

- Require the logarithm input to be positive. - Solve the resulting quadratic inequality. - What must a logarithm input equal for the logarithm to be \(0\)? - Verify that each solution lies in the domain.

1. The logarithm input must be positive: \(x^2-15>0\). Thus, \(x^2>15\), or \(|x|>\sqrt{15}\). The domain is \((-\infty, -\sqrt{15})\cup(\sqrt{15}, \infty)\). 2. For a zero, \(\ln(x^2-15)=0\), so \(x^2-15=1\). 3. Therefore, \(x^2=16\), giving \(x=-4\) and \(x=4\). Both values are in the domain.

Domain: \((-\infty, -\sqrt{15})\cup(\sqrt{15}, \infty)\) Zeros: \(x=-4\) and \(x=4\)

- Set the logarithm input greater than zero. - A logarithm equals zero when its input is one. - Check the solution against the domain.

1. The logarithm input must be positive: \(5-e^x>0\), so \(e^x<5\). Therefore, \(x<\ln(5)\), and the domain is \((-\infty, \ln(5))\). 2. For a zero, solve \(\ln(5-e^x)=0\). Then \(5-e^x=1\), so \(e^x=4\) and \(x=\ln(4)\). 3. Since \(\ln(4)<\ln(5)\), the zero is in the domain.

a) \((-\infty, \ln(5))\) b) \(x=\ln(4)\)

- Rewrite each logarithmic equation in exponential form. - Use roots or reciprocal powers to solve for the base. - Enforce the positive-base condition.

1. a) \(b^3=125\), so \(b=5\). 2. b) \(b^{-2}=0.01=\frac{1}{100}\), so \(b^2=100\). Since \(b>0\), \(b=10\). 3. c) \(b^{-3}=\frac{1}{8}\), so \(b^3=8\) and \(b=2\). 4. d) \(b^{1/2}=7\), so \(b=49\).

a) \(b=5\) b) \(b=10\) c) \(b=2\) d) \(b=49\)

- Compare the domains of \(\ln(x^6)\) and \(\ln(x)\). - Use an absolute value when applying the power property to an even power. - Solve the resulting absolute-value equation.

1. The original expression \(\ln(x^6)\) is defined for every \(x\neq0\), including negative values. 2. The step \(\ln(x^6)=6\ln(x)\) assumes \(x>0\). The identity valid for all \(x\neq0\) is \(\ln(x^6)=6\ln|x|\). 3. Solve \(6\ln|x|=12\): \(\ln|x|=2\), so \(|x|=e^2\). 4. Therefore, \(x=-e^2\) or \(x=e^2\).

Priya's step assumes \(x>0\) and loses the negative solution. The complete solution set is \(\{-e^2,e^2\}\).

- Find when the logarithm input is positive. - Condense the right-hand side into one logarithm. - Remember both solutions of a squared equation. - Check the domain.

1. The logarithm input must be positive, so \((x-5)^2>0\), which gives the domain \(\mathbb{R}\setminus\{5\}\). 2. Since \(2\ln(3)=\ln(9)\), the equation becomes \((x-5)^2=9\). 3. Thus, \(x-5=3\) or \(x-5=-3\), giving \(x=8\) or \(x=2\). 4. Both values lie in the domain.

Domain: \(\mathbb{R}\setminus\{5\}\); solution set: \(\{2,8\}\)

- Determine the common domain first. - Use the product property of logarithms. - Solve the resulting quadratic equation. - Reject values outside the domain.

1. The domain requires \(x+6>0\) and \(x>0\), so \(x>0\). 2. Combine the logarithms: \(\ln(x(x+6))=\ln(27)\). 3. Therefore, \(x^2+6x=27\), or \(x^2+6x-27=0\). 4. Factoring gives \((x-3)(x+9)=0\), so \(x=3\) or \(x=-9\). 5. Only \(x=3\) is in the domain.

\(\{3\}\)

- Determine the domain before solving. - Use the exponential function to undo \(\ln\). - A denominator cannot be zero. - Check each result in the original equation.

1. a) Exponentiating gives \(\frac{1}{x}=e^4\), so \(x=e^{-4}\), which is positive and valid. 2. b) The domain requires \(x>0\) and \(\ln(x)+1\neq0\). Multiplying gives \(4=2(\ln(x)+1)\), so \(\ln(x)=1\) and \(x=e\). This value is valid. 3. c) Exponentiating gives \(\sqrt{x+1}=e\). Squaring gives \(x+1=e^2\), so \(x=e^2-1\), which satisfies the domain.

a) \(x=e^{-4}\) b) \(x=e\) c) \(x=e^2-1\)

- First apply the domain restriction for the logarithm. - What additional restriction comes from having the logarithm in a denominator? - When does a fraction equal zero? - Check the zero against every domain restriction.

1. The logarithm input must be positive: \(x-3>0\), so \(x>3\). 2. The denominator cannot equal zero. Since \(\ln(x-3)=0\) when \(x-3=1\), exclude \(x=4\). 3. Therefore, the domain is \((3, 4)\cup(4, \infty)\). 4. A fraction is zero when its numerator is zero and its denominator is defined. Setting \(x-5=0\) gives \(x=5\). 5. Since \(5\) is in the domain, it is the only zero.

Domain: \((3, 4)\cup(4, \infty)\) Zero: \(x=5\)

- The logarithm must be defined and the square-root input must be nonnegative. - Determine when a natural logarithm is at least zero. - Square both sides and compare logarithm inputs. - Include both roots of the resulting quadratic equation.

1. The radicand must satisfy \(\ln(x^2-3)\geq0\). This condition also ensures that the logarithm input is positive. 2. Since \(\ln(u)\geq0\) exactly when \(u\geq1\), \(x^2-3\geq1\). Thus, \(x^2\geq4\), so the domain is \((-\infty, -2]\cup[2, \infty)\). 3. Set \(\sqrt{\ln(x^2-3)}=\sqrt{\ln(13)}\). Squaring gives \(\ln(x^2-3)=\ln(13)\). 4. Therefore, \(x^2-3=13\), so \(x^2=16\) and \(x=-4\) or \(x=4\). Both values lie in the domain.

a) \((-\infty, -2]\cup[2, \infty)\) b) \(x=-4\) and \(x=4\)

- Substitute the point coordinates. - Rewrite the logarithmic equation in exponential form. - Express \(32\) as a power of \(2\).

1. Substitute the point: \(7.5=3\log_b(32)\). 2. Divide by \(3\): \(\log_b(32)=2.5=\frac{5}{2}\). 3. Rewrite in exponential form: \(b^{5/2}=32\). 4. Thus, \(b=32^{2/5}=(2^5)^{2/5}=4\).

\(b=4\)

- Rewrite the logarithmic equation in exponential form. - Solve the resulting quadratic equation. - Check that the logarithm input is positive.

1. Rewrite in exponential form: \(x^2+2x=2^3=8\). 2. Solve \(x^2+2x-8=0\): \((x+4)(x-2)=0\), so \(x=-4\) or \(x=2\). 3. For both values, the logarithm input equals \(8>0\), so both are valid.

\(\{-4,2\}\)

- Determine where every logarithm argument is positive before doing any algebra. - Look for a logarithm property that combines the two terms on the right. - Equal logarithms with the same base have equal arguments when both sides are defined. - Check the candidate against the original domain.

1. The logarithm arguments require \(x-1>0\) and \(5-x>0\), so the domain is \(1<x<5\). 2. Use the product property to write \(\ln(5-x)+\ln(2)=\ln(2(5-x))\). 3. Since the logarithms are equal on their domain, \(x-1=2(5-x)\). Thus, \(x-1=10-2x\), so \(3x=11\) and \(x=\frac{11}{3}\). 4. Because \(1<\frac{11}{3}<5\), both logarithm arguments are positive. Therefore, \(x=\frac{11}{3}\) is admissible.

a) \(1<x<5\) b) \(x=\frac{11}{3}\) c) \(\frac{11}{3}\in(1,5)\), so both logarithms are defined.

- Determine where every logarithm argument is positive before doing algebra. - Look for a logarithm property that combines the two terms on the left. - A negative logarithm value behaves differently when the base is between \(0\) and \(1\); use the exponential definition rather than a sign shortcut. - Check every algebraic candidate in the original domain.

1. Both logarithm arguments must be positive, so \(x-1>0\) and \(x-3>0\). Thus, the domain is \(x>3\). 2. Combine the logarithms: \(\log_{\frac{1}{2}}\big((x-1)(x-3)\big)=-3\). 3. Rewrite in exponential form: \((x-1)(x-3)=\left(\frac{1}{2}\right)^{-3}=8\). 4. Expanding gives \(x^2-4x+3=8\), so \(x^2-4x-5=0\). 5. Factor: \((x-5)(x+1)=0\), giving candidates \(x=5\) and \(x=-1\). 6. Only \(x=5\) satisfies \(x>3\).

Domain: \(x>3\) Solution: \(x=5\)

- A substitution can turn an expression involving \(\log x\) into a quadratic inequality. - Use the product property to combine a sum of logarithms. - State the domain before solving, especially when the argument contains a square. - Use a sign chart or the graph of a quadratic to solve the resulting polynomial inequality.

1. The domain requires \(x^2 > 0\), so \(x \ne 0\). Since the common logarithm is increasing, \(\log(x^2) \le 2\) gives \(x^2 \le 100\), or \(-10 \le x \le 10\). Excluding \(0\) gives \([-10, 0) \cup (0, 10]\). 2. The domain requires \(x > 0\). Let \(u = \log x\). Then \(u^2 - 3u + 2 \le 0\), or \((u - 1)(u - 2) \le 0\). Thus, \(1 \le u \le 2\). Substituting back gives \(1 \le \log x \le 2\), so \(10 \le x \le 100\). 3. The domain requires \(x > 9\). Use the product property: \(\log(x(x - 9)) > 1\). Since the logarithm is increasing, \(x(x - 9) > 10\), so \(x^2 - 9x - 10 > 0\). Factoring gives \((x - 10)(x + 1) > 0\), so \(x < -1\) or \(x > 10\). Intersecting with \(x > 9\) gives \(x > 10\).

1) \([-10, 0) \cup (0, 10]\) 2) \([10, 100]\) 3) \((10, \infty)\)