How do you solve for x in 5^x=4^(x+1)?

1 Answer
Apr 21, 2016

$x \approx 6.21$

Explanation:

First we'll take the $\log$ of both sides:
$\log \left({5}^{x}\right) = \log \left({4}^{x + 1}\right)$
Now there's a rule in logarithms which is: $\log \left({a}^{b}\right) = b \log \left(a\right)$, saying that you can move any exponents down and out of the $\log$ sign. Applying this:
$x \log 5 = \left(x + 1\right) \log 4$
Now just rearrange to get x on one side
$x \log 5 = x \log 4 + \log 4$
$x \log 5 - x \log 4 = \log 4$
$x \left(\log 5 - \log 4\right) = \log 4$
$x = \log \frac{4}{\log 5 - \log 4}$
And if you type that into your calculator you'll get:
$x \approx 6.21 \ldots$