# How do you solve ln(e^x + e^-x) = ln (10/3)?

Jul 27, 2016

$x = \ln 3 \cong 1.098 , \mathmr{and} , x = \ln \left(\frac{1}{3}\right) = - \ln 3 \cong - 1.098$.

#### Explanation:

$\ln \left({e}^{x} + {e}^{-} x\right) = \ln \left(\frac{10}{3}\right)$

Since, ln is a $1 - 1$ fun., we get, ${e}^{x} + {e}^{-} x = \frac{10}{3}$

$\Rightarrow {e}^{x} + \frac{1}{e} ^ x = \frac{10}{3} = \frac{3}{1} + \frac{1}{3}$

By inspection, we can say that, ${e}^{x} = 3 , \mathmr{and} , \frac{1}{3}$. But let us proceed mathematically.

${e}^{x} + \frac{1}{e} ^ x = \frac{10}{3} \Rightarrow \frac{{e}^{2 x} + 1}{e} ^ x = \frac{10}{3} \Rightarrow 3 {e}^{2 x} + 3 = 10 {e}^{x}$

$\Rightarrow 3 {e}^{2 x} - 10 {e}^{x} + 3 = 0$

$\Rightarrow \left({e}^{x} - 3\right) \left(3 {e}^{x} - 1\right) = 0$

$\Rightarrow {e}^{x} = 3 , \mathmr{and} , {e}^{x} = \frac{1}{3}$

$x = \ln 3 , \mathmr{and} , x = \ln \left(\frac{1}{3}\right)$

To find $\ln 3 , \ln \left(\frac{1}{3}\right)$, we use the Change of Base Rule for Log. to get,

$x = \ln 3 = {\log}_{10} \frac{3}{\log} _ \left(10\right) e = \frac{0.4771}{0.4343} \cong 1.098$, and,

$x = \ln \left(\frac{1}{3}\right) = \ln 1 - \ln 3 = 0 - \ln 3 \cong - 1.098$

: