# How do you solve 7-2e^x=5?

May 22, 2015

Using $\ln$ is a useful way!

First, isolate ${e}^{x}$ in one side:

${e}^{x} = \frac{5 - 7}{- 2}$
${e}^{x} = 1$

Now, apply $\ln$ on both sides, to remove $x$ from the exponent, following the logarithm property $\ln {a}^{n} = n \cdot \ln a$

$\ln {e}^{x} = \ln 1$ (use a calculator for $\ln = 1$)
$x \ln e = \ln 1$

Now we have two definitions of $\ln$:

$\ln e = 1$ (because ${e}^{1} = e$)
and $\ln 1 = 0$ (because ${e}^{0} = 1$)

Thus,

$x \cdot 1 = 0$
$x = 0$