If x-0 and log2x-log2-x-log2 -x-log2-x- -4 -then x-

The correct option is C 4Given log_2x+log_2(√(x))+log_2 (√(x))+log_2(√(x))+...∞ =4 ⇒ log_2[x.x^1/2.x^1/4.x^1/8...∞ ]=log_2[x^1+12+14+18+...∞ ]=log_2 ...

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Byju's Answer

Standard X

Mathematics

Sum of Infinite Terms of a GP

If x>0 and ...

Question

If $x > 0$ and $l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + . . . \infty = 4,$ then $x =$

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Solution

The correct option is C 4
Given $l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + . . . \infty = 4$
$\Rightarrow l o g_{2} [x . x^{1 / 2} . x^{1 / 4} . x^{1 / 8} . . . \infty] = l o g_{2} [x^{1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . \infty}] = l o g_{2} x^{\frac{1}{1 - (1 / 2)}} = l o g_{2} (x^{2}) = 4$

$∴ x^{2} = 2^{4} = 16 ∴ x = 4$

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If x>0 and log2x+log2(√x)+log2(4√x)+log2(8√x)+...∞=4,then x=

The correct option is C 4Given log2x+log2(√x)+log2(4√x)+log2(8√x)+...∞=4⇒log2[x.x1/2.x1/4.x1/8...∞]=log2[x1+12+14+18+...∞]=log2x11−(1/2)=log2(x2)=4∴x2=24=16∴x=4

If $x > 0$ and $l o g_{2} x + l o g_{2} (\sqrt{x}) + l o g_{2} (\sqrt[4]{x}) + l o g_{2} (\sqrt[8]{x}) + . . . \infty = 4,$ then $x =$