How do you solve #4^x - 2^(x+1)-15=0#?

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Answer 1 · 2018-04-18T18:30:05

# x=log_2 5~~2.322##.

Observe that, #4^x=(2^2)^x=(2^x)^2, and, 2^(x+1)=2^x*2^1#.

Therefore, the given eqn. is, #(2^x)^2-2*2^x-15=0#.

So, if we subst. #2^x=y#, the eqn. becomes,

#y^2-2y-15=0#.

# (y-5)(y+3)=0#.

#:. y=5, or, y=-3#.

Since, #y=2^x gt 0, y=-3# is not admissible.

#:. y=5 rArr 2^x=5#.

#:. x=log_2 5#.

Using the Change of Base Rule, #x=log_10 5/log_10 2#,

#=0.699/0.301~~2.322#