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How do is solve this logarithm equation?

$$11 \cdot \log_3x+7 \cdot \log_7x = 13+3 \cdot \log_4x$$

I know that I have to use the change of base formula, but I still can't figure out the equation.

Can someone help me?

Arturo Magidin
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sarina eevers
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  • Is that eleven times $\log_3x$, or 11 times $x$ times $\log_3 x$? Very different things... – Arturo Magidin Feb 15 '19 at 17:42
  • But the image in the original question was $11x\log_3x + \cdots$. Here's the image: https://i.stack.imgur.com/q0rTG.png – stressed out Feb 15 '19 at 17:46
  • Which is it? The image you posted has an $x$, not a product. Such an equation would be much more difficult to solve than the one you have currently, with $11\log_3x$ instead of $11x\log_3 x$. Please make sure you are asking about the correct equation you are expected to solve. – Arturo Magidin Feb 15 '19 at 18:07
  • With the x I meant a product. Sorry for the confusion. – sarina eevers Feb 15 '19 at 18:13

2 Answers2

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The change of base formula is $$ \log_ab=\frac{\log b}{\log a} $$ where the base in the right hand side is whatever you prefer. I assume $e$. The equation becomes $$ \frac{11}{\log3}\log x+\frac{7}{\log7}\log x=13+\frac{3}{\log4}\log x $$ which is a first degree equation in $\log x$.

egreg
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You may want to use the identity: $\log_{a}(x)=\frac{\log(x)}{\log(a)}$

Thus $11 \cdot \log_3x+7 \cdot \log_7x = 13+3 \cdot\log_4x$ becomes:

$\frac{11\log(x)}{\log(3)}+\frac{7\log(x)}{\log(7)}=13+3\frac{\log(x)}{\log(4)}$.

Then set $u=\log(x)$ and solve for $u$, and then find $x$.

Arturo Magidin
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Locally unskillful
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