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?
$\endgroup$ 42 Answers
$\begingroup$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 $\log x$.
$\endgroup$ $\begingroup$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 $x$.
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