# How do you solve 2^x=7^(x-1)?

May 30, 2016

$x = \log \frac{7}{\log \left(7\right) - \log \left(2\right)} \approx 1.5533$

#### Explanation:

Dividing both sides by ${7}^{x}$ we find:

$\frac{1}{7} = {2}^{x} / {7}^{x} = {\left(\frac{2}{7}\right)}^{x}$

Then taking logs (any base):

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

Divide both sides by $\log \left(\frac{2}{7}\right)$ and transpose to get:

$x = \log \frac{\frac{1}{7}}{\log} \left(\frac{2}{7}\right) = \frac{- \log \left(7\right)}{\log \left(2\right) - \log \left(7\right)} = \log \frac{7}{\log \left(7\right) - \log \left(2\right)} \approx 1.5533$

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Complex solutions

We can find all Complex solutions by introducing a multiplier ${e}^{2 k \pi i} = 1$ during our calculations and using natural logarithms for convenience:

$\frac{1}{7} = {\left(\frac{2}{7}\right)}^{x} \cdot {e}^{2 k \pi i}$

$\ln \left(\frac{1}{7}\right) = \ln \left({\left(\frac{2}{7}\right)}^{x} \cdot {e}^{2 k \pi i}\right) = x \ln \left(\frac{2}{7}\right) + 2 k \pi i$

$x \ln \left(\frac{2}{7}\right) = \ln \left(\frac{1}{7}\right) - 2 k \pi i$

$x = \frac{\ln \left(\frac{1}{7}\right) - 2 k \pi i}{\ln} \left(\frac{2}{7}\right) = \frac{- \ln \left(7\right) - 2 k \pi i}{\ln \left(2\right) - \ln \left(7\right)} = \frac{\ln \left(7\right) + 2 k \pi i}{\ln \left(7\right) - \ln \left(2\right)}$

where $k$ is any integer.