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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