# How do you solve log 2^(x-1)= log 5^(x-2)?

Feb 20, 2016

x = $\frac{\log \left(2\right) - 2 \log \left(5\right)}{\log \left(2\right) - \log \left(5\right)} \approx 2.75647$

#### Explanation:

$\log {2}^{x - 1} = \log {5}^{x - 2}$

These logarithms have the same base

$L o {g}_{a} y = L o {g}_{a} x$

${a}^{L o {g}_{a} x} = {a}^{L o {g}_{a} y} - - - - - - - - - 1$

Now ${a}^{{\log}_{a} n} = n$

Extend this logic to 1

We get

$x = y$

Now lets use this result

$L o {g}_{a} y = L o {g}_{a} x \iff x = y$

So we get

${2}^{x - 1} = {5}^{x - 2}$

${2}^{x - 1} = {5}^{x - 1 - 1}$

${2}^{x - 1} = \frac{{5}^{x - 1}}{5}$

${2}^{x - 1} / \left({5}^{x - 1}\right) = \frac{1}{5}$

${\left(\frac{2}{5}\right)}^{x - 1} = \frac{1}{5}$

0.4)^(x-1 ) = 0.2

$x - 1 = L o {g}_{0.4} 0.2$

$x - 1 = 1.75647 \ldots$

$x = 1.75647 \ldots + 1$

$x = 2.75647 \ldots$

If you need it perfect logs

x = $\frac{\log \left(2\right) - 2 \log \left(5\right)}{\log \left(2\right) - \log \left(5\right)}$