# (Exponential Equation) How do I find X?

## ${5}^{x} = {4}^{x + 1}$

Apr 13, 2017

$x \approx 6.21$.

#### Explanation:

Take the natural logarithm of both sides.

$\ln \left({5}^{x}\right) = \ln \left({4}^{x + 1}\right)$

Now use $\ln {a}^{n} = n \ln a$.

$x \ln 5 = \left(x + 1\right) \ln 4$

$x \ln 5 = x \ln 4 + \ln 4$

$x \ln 5 - x \ln 4 = \ln 4$

$x \left(\ln 5 - \ln 4\right) = \ln 4$

This can be simplified further using $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$.

$x \left(\ln \left(\frac{5}{4}\right)\right) = \ln 4$

$x = \frac{\ln 4}{\ln \left(\frac{5}{4}\right)}$

If you prefer an approximation, we can take $x \approx 6.21$.

Hopefully this helps!

Apr 13, 2017

I got: $x = \frac{\ln \left(4\right)}{\ln \left(5\right) - \ln \left(4\right)}$

#### Explanation:

Here we can try applying the natural log, $\ln$, on both sides and apply some properties of logs:
$\ln {\left(5\right)}^{x} = \ln {\left(4\right)}^{x + 1}$
then:
$x \ln \left(5\right) = \left(x + 1\right) \ln \left(4\right)$
rearrange:
$x \ln \left(5\right) = x \ln \left(4\right) + \ln \left(4\right)$
$x \ln \left(5\right) - x \ln \left(4\right) = \ln \left(4\right)$
$x \left[\ln \left(5\right) - \ln \left(4\right)\right] = \ln \left(4\right)$
$x = \frac{\ln \left(4\right)}{\ln \left(5\right) - \ln \left(4\right)}$
if you have a pocket calculator we can easily evaluate the natural log to get:
$x = 6.21256$