How do you solve $7 - 2 e^{x} = 5$ $7-2e^x=5$ ?

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Answer 1 · 2015-05-22T23:52:15

Using $ln$ $ln$ is a useful way!

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

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

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

${ln e}^{x} = ln 1$ $lne^x=ln1$ (use a calculator for $ln = 1$ $ln=1$ )
$x ln e = ln 1$ $xlne=ln1$

Now we have two definitions of $ln$ $ln$ :

$ln e = 1$ $lne=1$ (because $e^{1} = e$ $e^1=e$ )
and $ln 1 = 0$ $ln1=0$ (because $e^{0} = 1$ $e^0=1$ )

Thus,

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

How do you solve 7−2ex=57-2e^x=5?