How do you solve 5(1.034)^x-1998=13?

Jul 23, 2015

$x = \frac{\ln \left(402.2\right)}{\ln \left(1.034\right)} \setminus \approx 179.363$

Explanation:

First, rearrange to write it as $5 {\left(1.034\right)}^{x} = 13 + 1998 = 2011$.

Next, divide both sides by $5$ to get ${1.034}^{x} = \frac{2011}{5} = 402.2$.

After that, take a logarithm of both sides (it doesn't matter what base you use). I'll use base $e$: $\ln \left({1.034}^{x}\right) = \ln \left(402.2\right)$ or, by a property of logarithms, $x \cdot \ln \left(1.034\right) = \ln \left(402.2\right)$.

Hence, $x = \frac{\ln \left(402.2\right)}{\ln \left(1.034\right)} \setminus \approx 179.363$.

You should check that this works by substitution back into the original equation.