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The original problem I have says:

Rewrite $3\ln (x - 5) - 2\ln (x + 4) + 3\ln (x + 5) - 2\ln (x - 4)$ as a single logarithm using the rules for logarithms.

But the answer and explanation both use $\log_8$ and no $\ln$.

I'm having trouble with how they came up with the $\log_8$. A picture is included below:

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4 Answers

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Note: You can replace each $\ln$ with (the same) logarithm to any base, and the result will be identical except for the base. In other words, I could replace every $\ln$ with a $\log_8$ in the answer below and it would still be valid.

You start with this expression:

$$3\ln (x - 5) - 2\ln (x + 4) + 3\ln (x + 5) - 2\ln (x - 4)$$

First use the power rule on each term:

$$\ln [(x - 5)^3] - \ln [(x + 4)^2] + \ln [(x + 5)^3] - \ln [(x - 4)^2]$$

Then use the product and quotient rule:

$$\ln\frac{(x-5)^3(x+5)^3}{(x-4)^2(x+4)^2}.$$

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If you want to rewrite it into one logarithm then, using the power rule on each term of $3\ln (x - 5) - 2\ln (x + 4) + 3\ln (x + 5) - 2\ln (x - 4)$ gives

$$\begin{align*} \ln (x-5)^3 + \ln (x+5)^3 - (\ln (x+4)^2 + \ln (x-4)^2) & = \ln (x-5)^3(x+5)^3 - \ln (x+4)^2(x-4)^2 \\ & = \ln \frac{((x-5)(x+5))^3}{((x-4)(x+4))^2} \\ & = \ln \frac{(x^2 - 25)^3}{(x^2 - 16)^2} \end{align*}$$

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Glad to see this got reopened. Anyway, it appears as though the original question was asked with $\ln$ and the answer key uses $\log_8$ in the solution. Either the question should have been in terms of $\log_8$ or the solution should have been in terms of $\ln$. It's an error on the problem and answer's author's part.

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Let $3\ln (x - 5) - 2\ln (x + 4) + 3\ln (x + 5) - 2\ln (x - 4) = y$

$e^{3\ln (x - 5) - 2\ln (x + 4) + 3\ln (x + 5) - 2\ln (x - 4)} = e^y$

$\frac{(x-5)^3(x+5)^3}{(x+4)^2(x-4)^2}= e^y$

From which we could do the following if we were really weird.

$\log_8 \frac{(x-5)^3(x+5)^3}{(x+4)^2(x-4)^2} = \log_8 e^y = y*\log_8 e$

$y = \frac{\log_8 \frac{(x-5)^3(x+5)^3}{(x+4)^2(x-4)^2}}{\log_8 e}$

$= \ln \frac{(x-5)^3(x+5)^3}{(x+4)^2(x-4)^2}$

Which really makes me think you wrote the question wrong from the very beginning that that the were not $\ln$ in the book but $\log_8$s.

Can you scan the question from the book? Not just the answer?

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