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Examples of Solving Logarithmic Equations

Examples of Logarithmic Equations

Solve each equation. Give exact values.

$LaTeX: 7\ln x=28$ $7\ln x=28$
$LaTeX: \log_2\left(x^3-19\right)=3$ $\log_2\left(x^3-19\right)=3$
$LaTeX: \log\left(x+6\right)-\log\left(x+2\right)=\log x$ $\log\left(x+6\right)-\log\left(x+2\right)=\log x$
$LaTeX: \log_2\left[\left(3x-7\right)\left(x-4\right)\right]=3$ $\log_2\left[\left(3x-7\right)\left(x-4\right)\right]=3$
$LaTeX: \log\left(3x+2\right)+\log\left(x-1\right)=1$ $\log\left(3x+2\right)+\log\left(x-1\right)=1$
$LaTeX: \ln e^{\ln x}-\ln\left(x-3\right)=\ln2$ $\ln e^{\ln x}-\ln\left(x-3\right)=\ln2$

Solutions

$LaTeX: 7\ln x=28$ $7\ln x=28$
$LaTeX: \ln x=4$ $\ln x=4$
Write in exponential form.
$x=e^4$
The solution set is { $e^4$ }
$LaTeX: \log_2\left(x^3-19\right)=3$ $\log_2\left(x^3-19\right)=3$
Write in exponential form.
$LaTeX: x^3-19=2^3\\ x^{3}-19=8\\ x^{3} =27$ $x^3-19=2^3\\ x^{3}-19=8\\ x^{3} =27$
Take the cube root of both sides.
$LaTeX: x=\sqrt[3]{27}\\ x=3$ $x=\sqrt[3]{27}\\ x=3$
The solution set is {3}.
$LaTeX: \log\left(x+6\right)-\log\left(x+2\right)=\log x$ $\log\left(x+6\right)-\log\left(x+2\right)=\log x$
Use the quotient property of logarithms.
$LaTeX: log\frac{x+6}{x+2}=\log x$ $log\frac{x+6}{x+2}=\log x$
Use the useful property of logarithms.
$LaTeX: \frac{x+6}{x+2}=x\\ x+6=x(x+2)\\ x+6=x^{2} +2x\\ x^{2} +x-6=0\\ (x+3)(x-2)=0$ $\frac{x+6}{x+2}=x\\ x+6=x(x+2)\\ x+6=x^{2} +2x\\ x^{2} +x-6=0\\ (x+3)(x-2)=0$
$LaTeX: x+3=0 or x-2=0\\ x=-3\:or\:x=2$ $x+3=0 or x-2=0\\ x=-3\:or\:x=2$
The proposed negative solution, -3, is not in the domain of log x in the original equation, so the only valid solution is the positive number 2. The solution set is {2}.
NOTE: Recall that the domain of $LaTeX: y=log_{a} x$ $y=log_{a} x$ is $LaTeX: \left(0,\infty\right)$ $\left(0,\infty\right)$ . For this reason, it is always necessary to check that proposed solutions of a logarithmic equation result in logarithms of positive numbers in the original equation.
$LaTeX: \log_2\left[\left(3x-7\right)\left(x-4\right)\right]=3$ $\log_2\left[\left(3x-7\right)\left(x-4\right)\right]=3$
$LaTeX: \left(3x-7\right)\left(x-4\right)=2^3\\ 3x^{2}-19x+28=8\\ 3x^{2}-19x+20=0\\ (3x-4)(x-5)=0\\ 3x-4=0\:or\:x-5=0\\ x=\frac{4}{3} \:or\:x=5$ $\left(3x-7\right)\left(x-4\right)=2^3\\ 3x^{2}-19x+28=8\\ 3x^{2}-19x+20=0\\ (3x-4)(x-5)=0\\ 3x-4=0\:or\:x-5=0\\ x=\frac{4}{3} \:or\:x=5$
A check is necessary to be sure that the argument of the logarithm in the given equation is positive. In both cases, the product $LaTeX: \left(3x-7\right)\left(x-4\right)$ $\left(3x-7\right)\left(x-4\right)$ is 8, and $LaTeX: \log_28=3$ $\log_28=3$ is true. The solution set is { $LaTeX: \frac{4}{3},5$ $\frac{4}{3},5$ }.
$LaTeX: \log\left(3x+2\right)+\log\left(x-1\right)=1$ $\log\left(3x+2\right)+\log\left(x-1\right)=1$
$LaTeX: \log_{10}\left[\left(3x+2\right)\left(x-1\right)\right]=1\\ (3x+2)(x-1)=10^{1} \\ 3x^{2} -x-2=10\\ 3x^{2}-x-12=0$ $\log_{10}\left[\left(3x+2\right)\left(x-1\right)\right]=1\\ (3x+2)(x-1)=10^{1} \\ 3x^{2} -x-2=10\\ 3x^{2}-x-12=0$
Using the quadratic formula we get:
$LaTeX: x=\frac{-\left(-1\right)\pm\sqrt[]{\left(-1\right)^2-4\left(3\right)\left(-12\right)}}{2\left(3\right)}=\frac{-1\pm\sqrt[]{145}}{6}$ $x=\frac{-\left(-1\right)\pm\sqrt[]{\left(-1\right)^2-4\left(3\right)\left(-12\right)}}{2\left(3\right)}=\frac{-1\pm\sqrt[]{145}}{6}$
The solution $LaTeX: \frac{-1-\sqrt[]{145}}{6}\:$ $\frac{-1-\sqrt[]{145}}{6}\:$ is negative, and when substituted for x in $LaTeX: lo\left(x-1\right)\:$ $lo\left(x-1\right)\:$ results in a negative argument, which is not allowed. Therefore this solution must be rejected. The solution set is { $LaTeX: \frac{-1+\sqrt[]{145}}{6}$ $\frac{-1+\sqrt[]{145}}{6}$ }.
$LaTeX: \ln e^{\ln x}-\ln\left(x-3\right)=\ln2$ $\ln e^{\ln x}-\ln\left(x-3\right)=\ln2$
Because $LaTeX: e^{\ln x}=x,\:$ $e^{\ln x}=x,\:$ the equation becomes:
$LaTeX: \ln x-\ln\left(x-3\right)=\ln2$ $\ln x-\ln\left(x-3\right)=\ln2$
$LaTeX: \ln\frac{x}{x-3}=\ln2\\ \frac{x}{x-3} =2\\ x=2x-6\\ 6=x$ $\ln\frac{x}{x-3}=\ln2\\ \frac{x}{x-3} =2\\ x=2x-6\\ 6=x$
Check that the solution set is {6}.