How do I solve this exponential equation?

Question

I have been trying to use substitution but that doesn't work unless its #4x^2#.

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Answer 1 · 2018-06-27T11:29:28

# x=4, or, x=3#.

Let, #2^x=y." Then, "4^x=(2^2)^x=(2^x)^2=y^2#.

Also, #2^(x+3)=2^x*2^3=8y#.

Thus, the eqn. becomes, #y^2-3*8y+128=0, #

# i.e., y^2-24y+128=0#.

#:. (y-16)(y-8)=0#.

#:. y=16, or, y=8#

#:. 2^x=16=2^4, or, y=8=2^3#.

#:. x=4, or, x=3#.