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

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Solution

The correct option is B 1
$A s s u m i n g b a s e a \frac{{log}_{a} a + {log}_{a} b c}{{log}_{a} a (1 + {log}_{a} b c)} = \frac{{log}_{a} a + {log}_{a} b c}{{log}_{a} a ({log}_{a} a + {log}_{a} b c)} = \frac{{log}_{a} a b c}{1 ({log}_{a} a b c)} = 1$

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Similar questions

\frac{l o g a + l o g b c}{l o g a (1 + l o g_{a} b c)}

S_{n} (x) = {log}_{a^{1 / 2}} x + {log}_{a^{1 / 3}} x + {log}_{a^{1 / 6}} x + {log}_{a^{1 / 11}} x + {log}_{a^{1 / 18}} x + {log}_{a^{1 / 27}} x + \dots up to n -terms,

a > 1.

S_{24} (x) = 1093

S_{12} (2 x) = 265,

a

f (x) = \frac{l o g (1 + a x) - l o g (1 - b x)}{x}

f (x) = \frac{l o g (1 + a x) - l o g (1 - b x)}{x}

f (x) = \frac{l o g (1 + a x) - l o g (1 - b x)}{x}

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