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Fundamental Laws of Logarithms

log logab a+1...

Question

$l o g {l o g_{a b} a + \frac{1}{l o g_{b} a b}} =$

A

\N

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B

1

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C

log ab

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D

None of these

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Solution

The correct option is A \N
$l o g {l o g_{a b} a + \frac{1}{l o g_{b} a b}} =$
= $l o g {l o g_{a b} a + l o g_{a b} b}$
= $l o g (l o g_{a b} a b) = l o g 1 = 0$

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2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}

= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}

a > b

a^{b} b^{b} > a^{b} b^{a}

log \frac{b}{a} < log \frac{1 + b}{1 + a}

x = 1 + {log}_{a} b c, y = 1 + {log}_{b} c a, z = 1 + {log}_{c} a b

x y + y z + z x = x y z

l o g_{a} (a b) = x

l o g_{b} (a b)

{log}_{a} b c = x, {log}_{b} c a = y, {log}_{c} a b = z

\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}

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Fundamental Laws of Logarithms

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