Question #3c7cf

Jun 16, 2017

See the explanation below.

Explanation:

I'm not sure if this is the exact question, I added a minus sign. But this is a similar question.

We will apply the following log law twice to solve for x:

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

First, see if you can simplify large numbers and write them in index form:

${\log}_{3} x - {\log}_{3} \left(\frac{x}{81}\right) = {\log}_{3} \left(\frac{x}{729}\right)$

$\Rightarrow {\log}_{3} x - {\log}_{3} \left(\frac{x}{3} ^ 4\right) = {\log}_{3} \left(\frac{x}{3} ^ 6\right)$

Apply the log law to the left hand side:

$\Rightarrow {\log}_{3} \left(\frac{\cancel{x}}{\frac{\cancel{x}}{3} ^ 4}\right) = {\log}_{3} \left(\frac{x}{3} ^ 6\right)$

$\Rightarrow {\log}_{3} \left({3}^{4}\right) = {\log}_{3} \left(\frac{x}{3} ^ 6\right)$

Bring both terms to the same side and apply the log law again:

$\Rightarrow {\log}_{3} \left({3}^{4}\right) - {\log}_{3} \left(\frac{x}{3} ^ 6\right) = 0$

$\Rightarrow {\log}_{3} \left(\frac{{3}^{4}}{\frac{x}{3} ^ 6}\right) = {\log}_{3} \left(\frac{{3}^{4} \cdot {3}^{6}}{x}\right) = {\log}_{3} \left(\frac{{3}^{10}}{x}\right) = 0$

Write the log expression in it's equivalent exponential form and then solve for x:

${\log}_{3} \left(\frac{{3}^{10}}{x}\right) = 0 \Rightarrow {3}^{0} = \frac{{3}^{10}}{x} = 1$

$\Rightarrow x = {3}^{10}$