Jason S. answered • 10/26/13

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Subtract 3 from both sides:

4

^{x}-10(4^{-x}) -3 = 0I multiplied through by 4

^{x}.4

^{x}(4^{x}-10*4^{-x}-3) = 4^{x}*04

^{2x}- 10* 4^{x-x}- 3*4^{x}= 0Since 4

^{x-x}= 4^{0 }= 1, rearrange:4

^{2x}-3(4^{x}) - 10 = 04

^{2x}is the same as 4^{(x)(2)}(4

^{x})^{2}-3(4^{x}) - 10 = 0.This is like a quadratic. Set u = 4

^{x }and substitute.u2 - 3u - 10 = 0

Factor:

(u-5)(u+2) = 0

Put the 4

^{x }back in for u:4

^{x}- 5 = 0 4^{x}+ 2 = 04

^{x}= 5 4^{x}= -2ln 4

^{x}= ln 5 No solution since there exists no power to which a positive 4 can be raised to get -2.x ln 4 = ln 5

x = ln 5/ln 4

Also:

ln 4 = ln 2

^{2 }= 2 ln 2