a) if f(x)=log(x+2), solve the equation: 2^(f(x−2)) ×2^(f(2x+2)) =4^(logf(x))


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Question Number 50080 by F_Nongue last updated on 13/Dec/18

$a) i f f (x) = l o g (x + 2), s o l v e t h e e q u a t i o n :$ $2^{f (x - 2)} \times 2^{f (2 x + 2)} = 4^{l o g f (x)}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18
	$f(x−2)=log(x−2+2)=logx f(2x+2)=log(2x+2+2)=log(2x+4) logf(x)=log{log(x+2)+2} 2^(logx) ×2^(log(2x+4)) =2^(2log{log(x+2)+2}) logx+log(2x+4)=2log{log(x+2)+2} log{x(2x+4)}=log{log(x+2)+2}^2 2x^2 +4={log(x+2)+2}^2 2x^2 +4=4+4log(x+2)+{log(x+2)}^2 2x^2 =4log(x+2)+{log(x+2)}^2 k^2 +4k−2x^2 =0 k=log(x+2) k^2 +4k+4=2x^2 +4 (k+2)^2 =2(x^2 +2) k+2=(√(2(x^2 +2))) log(x+2)+2=(√(2(x^2 +2)))$
	$f (x - 2) = l o g (x - 2 + 2) = l o g x$ $f (2 x + 2) = l o g (2 x + 2 + 2) = l o g (2 x + 4)$ $l o g f (x) = l o g {l o g (x + 2) + 2}$ $2^{l o g x} \times 2^{l o g (2 x + 4)} = 2^{2 l o g {l o g (x + 2) + 2}}$ $l o g x + l o g (2 x + 4) = 2 l o g {l o g (x + 2) + 2}$ $l o g {x (2 x + 4)} = l o g {l o g (x + 2) + 2}^{2}$ $2 x^{2} + 4 = {l o g (x + 2) + 2}^{2}$ $2 x^{2} + 4 = 4 + 4 l o g (x + 2) + {l o g (x + 2)}^{2}$ $2 x^{2} = 4 l o g (x + 2) + {l o g (x + 2)}^{2}$ $k^{2} + 4 k - 2 x^{2} = 0 k = l o g (x + 2)$ $k^{2} + 4 k + 4 = 2 x^{2} + 4$ ${(k + 2)}^{2} = 2 (x^{2} + 2)$ $k + 2 = \sqrt{2 (x^{2} + 2)}$ $l o g (x + 2) + 2 = \sqrt{2 (x^{2} + 2)}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18

	Commented by Pk1167156@gmail.com last updated on 14/Dec/18
	nice sir
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