How do you simplify log_7 9x+log_7x-3log_7 x?

Dec 12, 2015

${\log}_{7} \left(\frac{9}{x}\right)$

Explanation:

First of all, observe that $3 {\log}_{7} \left(x\right) = {\log}_{7} \left({x}^{3}\right)$. So, your expression becomes

${\log}_{7} \left(9 x\right) + {\log}_{7} \left(x\right) - {\log}_{7} \left({x}^{3}\right)$

Now, the sum of two logarithms is the logarithm of the product:

${\log}_{7} \left(a\right) + {\log}_{7} \left(b\right) = {\log}_{7} \left(a b\right)$

So,

$\textcolor{g r e e n}{{\log}_{7} \left(9 x\right) + {\log}_{7} \left(x\right)} - {\log}_{7} \left({x}^{3}\right) = \textcolor{g r e e n}{{\log}_{7} \left(9 {x}^{2}\right)} - {\log}_{7} \left({x}^{3}\right)$

And the difference of two logarithms is the logarithm of the ratio:

${\log}_{7} \left(a\right) - {\log}_{7} \left(b\right) = {\log}_{7} \left(\frac{a}{b}\right)$

So,

${\log}_{7} \left(9 {x}^{2}\right) - {\log}_{7} \left({x}^{3}\right) = {\log}_{7} \left(9 {x}^{2} / {x}^{3}\right) = {\log}_{7} \left(\frac{9}{x}\right)$