-log2x-0-5-log2-x- then x equals

√(log_2x) 0.5=log_2√(x) (log_2x)^1/2=1/2+1/2log_2x log_2x=1/4+1/4(log_2x)^2+1/2log_2x (log_2x)^2 2log_2x+1=0 (log_2x 1)^2=0 log_2x=1 x=2 ...

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Fundamental Laws of Logarithms

√log2x-0.5=lo...

Question

$\sqrt{l o g_{2} x} - 0.5 = l o g_{2} \sqrt{x}$ , then x equals

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\sqrt{{log}_{2} x} - 0.5 = {log}_{2} \sqrt{x}

x

\sqrt{{log}_{2} x} - 0.5 = {log}_{2} \sqrt{x}

x

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}) + l o g_{2} (\sqrt[16]{x}) + . . . . . = 4

$\sqrt{l o g_{2} x} - 0.5 = l o g_{2} \sqrt{x}$ , then x equals

\sqrt{{log}_{2} x} - 0.5 = {log}_{2} \sqrt{x}

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