How do you solve ln(e^x + e^-x) = ln (10/3)ln(ex+e−x)=ln(103)?

1 Answer
Ratnaker Mehta
Jul 27, 2016

x=ln3~=1.098, or, x=ln(1/3)=-ln3~=-1.098x=ln3≅1.098,or,x=ln(13)=−ln3≅−1.098.

Explanation:

ln(e^x+e^-x)=ln(10/3)ln(ex+e−x)=ln(103)

Since, ln is a 1-11−1 fun., we get, e^x+e^-x=10/3ex+e−x=103

rArre^x+1/e^x=10/3=3/1+ 1/3⇒ex+1ex=103=31+13

By inspection, we can say that, e^x=3, or, 1/3ex=3,or,13. But let us proceed mathematically.

e^x+1/e^x=10/3rArr(e^(2x)+1)/e^x=10/3rArr3e^(2x)+3=10e^xex+1ex=103⇒e2x+1ex=103⇒3e2x+3=10ex

rArr3e^(2x)-10e^x+3=0⇒3e2x−10ex+3=0

rArr (e^x-3)(3e^x-1)=0⇒(ex−3)(3ex−1)=0

rArre^x=3, or, e^x=1/3⇒ex=3,or,ex=13

x=ln3, or, x=ln(1/3)x=ln3,or,x=ln(13)

To find ln3, ln(1/3)ln3,ln(13), we use the Change of Base Rule for Log. to get,

x=ln3=log_(10)3/log_(10)e=0.4771/0.4343~=1.098x=ln3=log103log10e=0.47710.4343≅1.098, and,

x=ln(1/3)=ln1-ln3=0-ln3~=-1.098x=ln(13)=ln1−ln3=0−ln3≅−1.098

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