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Solve the following for $x$ . $\newline$ $\log_{3}(x^{2}-24)=\log_{3}(5x)$

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Quotient property of logarithms

Full solution

Q. Solve the following for

x

.

\newline

\log_{3}(x^{2}-24)=\log_{3}(5x)

Set Equations Equal: Set the insides of the logarithms equal to each other. $\newline$ Since the bases of the logarithms are the same, we can set the arguments (the insides of the logarithms) equal to each other. $\newline$ $x^2 - 24 = 5x$
Rearrange to Quadratic: Rearrange the equation to form a quadratic equation. $\newline$ Subtract $5x$ from both sides to get all terms on one side of the equation. $\newline$ $x^2 - 5x - 24 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $-24$ and add to $-5$ . These numbers are $-8$ and $3$ . $\newline$ $(x - 8)(x + 3) = 0$
Solve Using Zero Product Property: Solve for $x$ using the zero product property. $\newline$ Set each factor equal to zero and solve for $x$ . $\newline$ $x - 8 = 0$ or $x + 3 = 0$ $\newline$ $x = 8$ or $x = -3$
Check for Extraneous Solutions: Check for extraneous solutions. $\newline$ We must check the solutions in the original logarithmic equation to ensure they do not make the argument of the logarithm negative or zero, as the logarithm is not defined for non-positive values. $\newline$ For $x = 8$ : $\newline$ $\log_3(8^2 - 24) = \log_3(5\times8)$ $\newline$ $\log_3(64 - 24) = \log_3(40)$ $\newline$ $\log_3(40) = \log_3(40)$ $\newline$ This is a valid solution. $\newline$ For $x = -3$ : $\newline$ $\log_3((-3)^2 - 24) = \log_3(5\times(-3))$ $\newline$ $\log_3(9 - 24) = \log_3(-15)$ $\newline$ $\log_3(-15)$ is not defined, so $x = -3$ is an extraneous solution.

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