Question

log m+n/7= 1/2( log m+ log n). m/n+n/m=47
​

Answer 1

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

Given that

$\rm :\longmapsto\:log\bigg[\dfrac{m + n}{7} \bigg] = \dfrac{1}{2} \bigg(logm +logn \bigg)$

can be rewritten as

$\rm :\longmapsto\:2 \: log\bigg[\dfrac{m + n}{7} \bigg] = logm +logn$

We know that

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

and

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

So, using this, we get

$\rm :\longmapsto\:log {\bigg[\dfrac{m + n}{7} \bigg]}^{2} = logmn$

$\rm :\longmapsto\:{\bigg[\dfrac{m + n}{7} \bigg]}^{2} = mn$

$\rm :\longmapsto\:\dfrac{ {m}^{2} + {n}^{2} + 2mn}{49} = mn$

$\rm :\longmapsto\: {m}^{2} + {n}^{2} + 2mn = 49mn$

$\rm :\longmapsto\: {m}^{2} + {n}^{2}= 49mn - 2mn$

$\rm :\longmapsto\: {m}^{2} + {n}^{2}= 47mn$

$\rm :\longmapsto\:\dfrac{ {m}^{2} + {n}^{2} }{mn} = 47$

$\rm :\longmapsto\:\dfrac{ {m}^{2} }{mn} + \dfrac{ {n}^{2} }{mn} = 47$

$\rm :\longmapsto\:\dfrac{m}{n} + \dfrac{n}{m} = 47$

Hence, Proved

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Explore More

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

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

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

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

$\boxed{\tt{ log_{ {x}^{n} }( {x}^{m} ) = \frac{m}{n} \: }}$

$\boxed{\tt{ log_{ {x}^{n} }( {y}^{m} ) = \frac{m}{n} log_{x}(y) \: }}$

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

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

$\boxed{\tt{ {e}^{logx} = x \: }}$

Answer 2

Answer:

$\mathtt{ \implies \: log( \frac{m + n}{7} ) = log( \sqrt{mn} ) }$

$\mathtt{taking \: antilog \: \frac{m + n}{7} = \sqrt{mn} }$

$\mathtt{on \: sqaring \: {m + n}^{2} = (7 \sqrt{ mn} )^{2} }$

$\mathtt{ \implies \: {m}^{2} + {n}^{2} + 2mn = 49mn }$

$\mathtt{ \implies \: \frac{ {m}^{2} + {n}^{2} + 2mn }{mn} = 49}$

$\mathtt{ \implies \: \frac{m}{n} + \frac{n}{n} + 2 = 49 }$

$\mathtt{ \implies \: \frac{m}{n} + \frac{n}{m} = 47 }$

$\mathtt{ \: hence \: proved}$