Rewrite \$3ln (x - 5) - 2ln (x + 4) + 3ln (x + 5) - 2ln (x - 4)\$ together a solitary logarithm making use of the rules for logarithms.

You are watching: Ln (x / 5) = 2

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

I"m having actually trouble with just how they come up v the \$log_8\$. A snapshot is included below:

\$egingroup\$ uncover the base? The base is \$e\$ due to the fact that the logs in the expression space \$ln\$. Is that what you're asking? \$endgroup\$
\$egingroup\$ climate there is essential information absent from this question. As it's written appropriate now, the prize is \$e\$. \$endgroup\$
\$egingroup\$ I see you've modified the concern text, which significantly changes the problem. Response involving \$log_8\$ tho doesn't do sense, since if every you're law is rewriting the provided expression together a single logarithm, then there's no reason to readjust the base. If the answer vital really has actually \$log_8\$ for that exact trouble you've written, climate the answer vital is incorrect. \$endgroup\$
Note: You deserve to replace every \$ln\$ with (the same) logarithm to any kind of base, and the result will be identical other than for the base. In various other words, I can replace every \$ln\$ with a \$log_8\$ in the price below and it would certainly still be valid.

You begin with this expression:

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

First use the power preeminence on each term:

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

Then usage the product and quotient rule:

\$\$lnfrac(x-5)^3(x+5)^3(x-4)^2(x+4)^2.\$\$

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edited Aug 2 "16 in ~ 19:51
answer Aug 1 "16 in ~ 20:35

JohnJohn
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If you desire to rewrite it right into one logarithm then, making use of the power dominion on each term of \$3ln (x - 5) - 2ln (x + 4) + 3ln (x + 5) - 2ln (x - 4)\$ offers

\$\$eginalign* 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 endalign*\$\$

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answered Aug 1 "16 at 20:35

Zain PatelZain Patel
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Glad to view this got reopened. Anyway, it shows up as though the initial question to be asked v \$ln\$ and also the answer key uses \$log_8\$ in the solution. Either the question need to have been in terms the \$log_8\$ or the solution should have remained in terms that \$ln\$. It"s an error ~ above the problem and answer"s author"s part.

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answered Aug 2 "16 at 15:30
user307169user307169
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Let \$3ln (x - 5) - 2ln (x + 4) + 3ln (x + 5) - 2ln (x - 4) = y\$

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

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

From which us could perform the complying with if we were yes, 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 = fraclog_8 frac(x-5)^3(x+5)^3(x+4)^2(x-4)^2log_8 e\$

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

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

Can girlfriend scan the question indigenous the book? Not just the answer?

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answer Aug 2 "16 at 20:09

fleabloodfleablood
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