Question

1/(log4 6) + 1/(log9 6)​

Answer 1

$\large\underline{\sf{Solution-}}$

The given expression is

$\rm \: \dfrac{1}{ log_{4}(6) } + \dfrac{1}{ log_{9}(6) }$

We know,

$\boxed{\tt{ \: \: log_{x}(y) = \frac{1}{ log_{y}(x) } \: \: }}$

So, using this result, given expression can be rewritten as

$\rm \: = \: log_{6}(4) + log_{6}(9)$

We know

$\boxed{\tt{ \: log_{m}(x) + log_{m}(y) \: = \: log_{m}(xy) \: }}$

So, using this result, we get

$\rm \: = \: log_{6}(9 \times 4)$

$\rm \: = \: log_{6}(36)$

$\rm \: = \: log_{6}(6 \times 6)$

$\rm \: = \: log_{6}( {6}^{2} )$

We know

$\boxed{\tt{ \: log_{a}( {a}^{x} ) = x \: }}$

So, using this result, we get

$\rm \: = \: 2$

Hence,

$\rm\implies \:\boxed{\tt{ \: \rm \: \dfrac{1}{ log_{4}(6) } + \dfrac{1}{ log_{9}(6) } = 2 \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ logx - logy = log \frac{x}{y} \: }}$

$\boxed{\tt{ log {x}^{y} = y \: logx \: }}$

$\boxed{\tt{ log_{x}(x) = 1 \: }}$

$\boxed{\tt{ log_{ {x}^{a} }( {x}^{b} ) = \frac{b}{a} \: }}$

$\boxed{\tt{ log_{ {x}^{a} }( {y}^{b} ) = \frac{b}{a} log_{x}(y) \: }}$

$\boxed{\tt{ {a}^{ log_{a}(x) } = x \: }}$

$\boxed{\tt{ {a}^{ ylog_{a}(x) } = {x}^{y} \: }}$

$\boxed{\tt{ {x}^{y} = z \: \: \rm\implies \:y = log_{x}(z) \: }}$

$\boxed{\tt{ \: log1 = 0 \: }}$