# How do you solve 12^(2x-3)=16?

Jul 4, 2015

This suggests that you can use ${\log}_{12} \left(x\right)$, where $x$ is either ${12}^{2 x - 3}$ or $16$. (By default, $\log x = {\log}_{10} x$.)

${\log}_{12} \left({12}^{2 x - 3}\right) = {\log}_{12} \left(16\right)$

Now if you use the change of base law, you get:

$\frac{\log \left({12}^{2 x - 3}\right)}{\log 12} = \frac{\log 16}{\log 12}$

Since $\frac{\ln x}{\log x} = \text{constant}$ (try it on any $x$; it is $\approx 2.303$), we can change this to:

$\frac{\ln \left({12}^{2 x - 3}\right)}{\ln 12} = \frac{\ln 16}{\ln 12}$

Using exponential rules of logs:

$\left(2 x - 3\right) \cancel{\left(\frac{\ln 12}{\ln 12}\right)} = \frac{\ln 16}{\ln 12}$

$2 x - 3 = \frac{\ln 16}{\ln 12}$

$x = \textcolor{b l u e}{\frac{\frac{\ln 16}{\ln 12} + 3}{2} \approx 2.05789}$

If you plug this back in, it indeed works.

${12}^{\left(2 \left(\approx 2.05789\right) - 3\right)} = 16$

${12}^{\left(\approx 1.11578\right)} = 16$