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How do you solve four to the x power minus two to the x plus one power all equaling 15?
Wiki User
∙ 12y ago
So you have (4^(x-2))^(x+1)=15. Using a law of exponents (b^a)^c=b^(a*c) and 4=2^2, you get
4^((x-2)*(x+1))=15=2^(2(x-2)(x+1)).
Take log base two and use the rule about logs of products you get
2(x-2)(x+1)=Log_2(15)
Simplifying and substituting u for Log_2(15),
2x^2-2x-4=u, which means
2x^2-2x-(4+u)=0, using quadratic formula.
X=(1/4)(2(+or-)sqrt(4-4*2*(-4-u))) or
(1/2) plus or minus (1/4)(sqrt(36+8u))
Where sqrt(b) is the square root of b
Wiki User
∙ 12y ago
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