# How do you solve log_3 (x) = 9log_x (3)?

Mar 20, 2016

$x = \frac{1}{27}$ or $27$

#### Explanation:

Use the change of base formula:

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

Here, we will use the arbitrary base $c = 10$, so we have

${\log}_{a} \left(b\right) = {\log}_{10} \frac{b}{\log} _ 10 \left(a\right) = \log \frac{b}{\log} \left(a\right)$

Recall that ${\log}_{10} \left(x\right) = \log \left(x\right)$, as the common logarithm.

Applying the change of base formula, we see that

$\log \frac{x}{\log} \left(3\right) = \frac{9 \log \left(3\right)}{\log} \left(x\right)$

Cross multiply.

${\log}^{2} \left(x\right) = 9 {\log}^{2} \left(3\right)$

Take the square root of both sides. Don't forget that this can be positive or negative!

$\log \left(x\right) = \pm 3 \log \left(3\right)$

We can split this into the two equations:

$\left.\left(\log \left(x\right) = 3 \log \left(3\right) , \text{ "" or "" } , \log \left(x\right) = - 3 \log \left(3\right)\right)\right.$

Rewrite the right hand side of both equations using the rule:

$b \cdot \log \left(a\right) = \log \left({a}^{b}\right)$

This gives us

$\left.\begin{matrix}\log \left(x\right) = \log \left({3}^{3}\right) & \text{ "color(white)"or"" " & log(x)=log(3^-3) \\ log(x)=log(27) & " "color(white)"or"" " & log(x)=log(1/27) \\ " "" "x=27 & " "color(white)"or"" " & " "" } x = \frac{1}{27}\end{matrix}\right.$