# How to solve this logarithmic function?

Apr 26, 2018

$x = 1 , 2$

#### Explanation:

First, I would get both the logarithms on the same side.

${\log}_{2} \left({3}^{2 x - 2} + 7\right) - {\log}_{2} \left({3}^{x - 1} + 1\right) = 2$

Now, use the rule $\log \left(A\right) - \log \left(B\right) = \log \left(A / B\right)$:

${\log}_{2} \left(\frac{{3}^{2 x - 2} + 7}{{3}^{x - 1} + 1}\right) = 2$

Rewrite by undoing the logarithm:

$\frac{{3}^{2 x - 2} + 7}{{3}^{x - 1} + 1} = {2}^{2} = 4$

Cross-multiply:

${3}^{2 x - 2} + 7 = 4 \left({3}^{x - 1} + 1\right)$

${3}^{2 x - 2} + 7 = 4 \left({3}^{x - 1}\right) + 4$

${3}^{2 x - 2} + 3 = 4 \left({3}^{x - 1}\right)$

I stared at this for a little while before I realized something cool: that $2 x - 2 = 2 \left(x - 1\right)$. The relevance of this is that we can rewrite the equation as follows:

${\left({3}^{x - 1}\right)}^{2} - 4 \left({3}^{x - 1}\right) + 3 = 0$

Now, we have a quadratic equation, if we let $t = {3}^{x - 1}$.

${t}^{2} - 4 t + 3 = 0$

And solve as normal:

$\left(t - 1\right) \left(t - 3\right) = 0$

$t = 1 , 3$

Now we have to solve for $x$ using $t = {3}^{x - 1}$. In the case $t = 1$:

${3}^{x - 1} = 1$

Take the logarithm:

$x - 1 = {\log}_{3} \left(1\right) = 0$

$x = 1$

We also need the case $t = 3$:

${3}^{x - 1} = 3$

$x - 1 = {\log}_{3} \left(3\right) = 1$

$x = 2$