# How do you solve 10^x=5^(x+1)?

May 7, 2016

$x = \log \frac{5}{\log} 2$.

#### Explanation:

${10}^{x} = {\left(2 X 5\right)}^{x} = {2}^{x} {5}^{x} = {5}^{x + 1} = {5}^{x} {5}^{1}$.

Cancelling the common factor ${5}^{x}$,

${2}^{x} = 5$.

Equating the logarithms,

$x \log 2 = \log 5$.

So, $x = \log \frac{5}{\log} 2$.

May 7, 2016

I found: $x = 2.32193$

#### Explanation:

We can try taking the natural log of both sides and apply one property of logs to write:
$\ln \left({10}^{x}\right) = \ln \left({5}^{x + 1}\right)$
and then for the exponents:
$x \ln \left(10\right) = \left(x + 1\right) \ln \left(5\right)$
rearrange:
$x \ln \left(10\right) = x \ln \left(5\right) + \ln \left(5\right)$
$x \ln \left(10\right) - x \ln \left(5\right) = \ln \left(5\right)$
so:
$x \left[\ln \left(10\right) - \ln \left(5\right)\right] = \ln \left(5\right)$
and:
$x = \frac{\ln \left(5\right)}{\ln \left(10\right) - \ln \left(5\right)} = \ln \frac{5}{\ln} \left(\frac{10}{5}\right) = \ln \frac{5}{\ln} 2 = 2.32193$