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

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Answer 1 · 2016-07-27T16:19:05

$x=ln3~=1.098, or, x=ln(1/3)=-ln3~=-1.098$ .

$ln(e^x+e^-x)=ln(10/3)$

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

$rArre^x+1/e^x=10/3=3/1+ 1/3$

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

$e^x+1/e^x=10/3rArr(e^(2x)+1)/e^x=10/3rArr3e^(2x)+3=10e^x$

$rArr3e^(2x)-10e^x+3=0$

$rArr (e^x-3)(3e^x-1)=0$

$rArre^x=3, or, e^x=1/3$

$x=ln3, or, x=ln(1/3)$

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

$x=ln3=log_(10)3/log_(10)e=0.4771/0.4343~=1.098$ , and,

$x=ln(1/3)=ln1-ln3=0-ln3~=-1.098$

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