# How do you solve lnx=7.25?

Jul 15, 2016

$x = {e}^{7.25} \approx 1408.105$

#### Explanation:

Remember from basic definitions
$\textcolor{w h i t e}{\text{XXX}} \ln \left(a\right) = c$ means ${e}^{c} = a$

Therefore if
$\textcolor{w h i t e}{\text{XXX}} \ln \left(x\right) = 7.25$
then
$\textcolor{w h i t e}{\text{XXX}} x = {e}^{7.25}$
(using a calculator we can find the approximation ${e}^{7.25} = 1408.105$)

Jul 15, 2016

${e}^{7.25}$, or ${10}^{7.25}$, or ${b}^{7.25}$, $b$ being the base of the logarithm.

#### Explanation:

You can only use the fact that the logarithm and the exponential are one the inverse function of the other. This means that

${e}^{\ln} x = x$ and $\ln \left({e}^{x}\right) = x$

Using this property, you can put both left and right member at the exponent:

$\ln \left(x\right) = 7.25 \setminus \iff {e}^{\ln} \left(x\right) = {e}^{7.25}$

But we have just observed that ${e}^{\ln} \left(x\right) = x$, so we have

$x = {e}^{7.25}$.

This, of course, assuming that by "log" you meant the natural one. If, for example, you use base 10 logarithm, you should change $e$ with $10$, like this:

$\log \left(x\right) = 7.25 \setminus \iff {10}^{\log} \left(x\right) = {10}^{7.25} \setminus \iff x = {10}^{7.25}$