# How do you solve 2.5 = ln(x)?

$x = {e}^{2.5} \approx 12.1825$
${\log}_{b} \left(c\right) = d$ is equivalent to ${b}^{d} = c$
Since $\ln \left(x\right) = {\log}_{e} \left(x\right)$, $\ln \left(x\right) = 2.5$ is equivalent to ${e}^{2.5} = x$