How do you solve 4^x + 6(4^-x) = 54x+6(4−x)=5?

1 Answer
Massimiliano
May 3, 2015

In this way:

Multiply both the members of the equation for 4^x4x, 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=04x⋅4x+6⋅4−x4x−5⋅4x=0⇒42x−5⋅4x+6=0.

This is a quadratric equation in the variable 4^x4x, 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.

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