Question

$\begin{gathered}\large {\boxed{\sf{\mid{\overline {\underline {\star QUESTION ::}}}\mid}}}\end{gathered}$


$if \: \: \frac{1}{ log_{a}(x) } + \frac{1}{ log_{b}(x) } = \frac{2}{ log_{c}(x) } \\ \\ prove \: \: that \: \: c {}^{2} = ab$
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Answer 1

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

$\rm :\longmapsto\:\dfrac{1}{ log_{a}(x) } + \dfrac{1}{ log_{b}(x) } = \dfrac{2}{ log_{c}(x) }$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: {c}^{2} = ab$

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

$\rm :\longmapsto\:\dfrac{1}{ log_{a}(x) } + \dfrac{1}{ log_{b}(x) } = \dfrac{2}{ log_{c}(x) }$

We know that,

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

So, using this identity, we have

$\rm :\longmapsto\: log_{x}(a) + log_{x}(b) = 2 log_{x}(c)$

Now, we know that

$\boxed{ \rm \: log_{x}(y) + log_{x}(z) = log_{x}(yz)}$

So, using this identity, we get

$\rm :\longmapsto\: log_{x}(ab)= 2 log_{x}(c)$

Now, again we know that

$\boxed{ \rm \: y log(x) = log( {x}^{y} )}$

Using this result, we get

$\rm :\longmapsto\: log_{x}(ab)= log_{x}( {c}^{2} )$

Now, we know,

$\boxed{ \rm \: log_{a}(x) = log_{a}(y) \: \implies \: x \: = \: y}$

So, using this, identity,

$\rm :\longmapsto\:ab = {c}^{2}$

$\bf\implies \: {c}^{2} = ab$

Hence, Proved

Additional Information :-

$\boxed{ \rm \: logx \: + \: logy \: = \: logxy}$

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

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

$\boxed{ \rm \: log_{x}(y) = \frac{logy}{logx} }$

$\boxed{ \rm \: log_{ {x}^{a} }( {y}^{a} ) = \frac{logy}{logx} = log_{x}(y) }$

Answer 2

Answer:

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