How do you solve #4^x + 6(4^-x) = 5#?

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Answer 1 · 2015-05-03T19:08:47

In this way:

Multiply both the members of the equation for #4^x#, that is always a positive (not zero) term.

#4^x*4^x+6*4^-x4^x-5*4^x=0rArr4^(2x)-5^*4^x+6=0#.

This is a quadratric equation in the variable #4^x#, so:

#Delta=b^2-4ac=25-24=1=1^2#,

#4^x=(-b+-sqrtDelta)/(2a)=(5+-1)/2rArr#

#4^x=((5+1)/2)=3rArrln4^x=ln3rArrxln4=ln3rArrx=ln3/ln4#;

#4^x=(5-1)/2=2rArr2^(2x)=2^1rArr2x=1rArrx=1/2#.