How do you solve $ln (e^{x} + e^{- x}) = ln (\frac{10}{3})$ $ln(e^x + e^-x) = ln (10/3)$ ?

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How do you solve $ln (e^{x} + e^{- x}) = ln (\frac{10}{3})$ $ln(e^x + e^-x) = ln (10/3)$ ?

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Answer 1 · 2016-07-27T16:19:05

$x = ln 3 ≅ 1.098, or, x = ln (\frac{1}{3}) = - ln 3 ≅ - 1.098$ $x=ln3~=1.098, or, x=ln(1/3)=-ln3~=-1.098$ .

Explanation:

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

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

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

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

$e^{x} + \frac{1}{e^{x}} = \frac{10}{3} \Rightarrow \frac{e^{2 x} + 1}{e^{x}} = \frac{10}{3} \Rightarrow 3 e^{2 x} + 3 = 10 e^{x}$ $e^x+1/e^x=10/3rArr(e^(2x)+1)/e^x=10/3rArr3e^(2x)+3=10e^x$

$\Rightarrow 3 e^{2 x} - 10 e^{x} + 3 = 0$ $rArr3e^(2x)-10e^x+3=0$

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

$\Rightarrow e^{x} = 3, or, e^{x} = \frac{1}{3}$ $rArre^x=3, or, e^x=1/3$

$x = ln 3, or, x = ln (\frac{1}{3})$ $x=ln3, or, x=ln(1/3)$

To find $ln 3, ln (\frac{1}{3})$ $ln3, ln(1/3)$ , we use the Change of Base Rule for Log. to get,

$x = ln 3 = \frac{{log}_{10} 3}{{log}_{10} e} = \frac{0.4771}{0.4343} ≅ 1.098$ $x=ln3=log_(10)3/log_(10)e=0.4771/0.4343~=1.098$ , and,

$x = ln (\frac{1}{3}) = ln 1 - ln 3 = 0 - ln 3 ≅ - 1.098$ $x=ln(1/3)=ln1-ln3=0-ln3~=-1.098$

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How do you solve $ln (e^{x} + e^{- x}) = ln (\frac{10}{3})$ $ln(e^x + e^-x) = ln (10/3)$ ?

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Explanation:

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