Question

Pls tell the answer fastt

Answer 1

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

Given that

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

and

$\rm :\longmapsto\: log_{5}(y) = a$

We know that,

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

So, using this identity,

$\rm :\longmapsto\:x = {2}^{a}$

and

$\rm :\longmapsto\:y = {5}^{a}$

Also, given that

$\rm :\longmapsto\: {100}^{2a - 1} = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

We know that,

$\red{\boxed{ \rm{ {a}^{x - y} = \frac{ {a}^{x} }{ {a}^{y} }}}}$

So, using this identity, we get

$\rm :\longmapsto\: \dfrac{ {100}^{2a} }{100} = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

We know,

$\red{\boxed{ \rm{ {( {a}^{x} )}^{y} = {a}^{xy}}}}$

$\rm :\longmapsto\: \dfrac{ {10}^{4a} }{ {10}^{2} } = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

$\rm :\longmapsto\: \dfrac{ {(2 \times 5)}^{4a} }{ {10}^{2} } = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

We know that,

$\red{\boxed{ \rm{ {(xy)}^{a} = {x}^{a} \times {y}^{a}}}}$

So, using this, we get

$\rm :\longmapsto\: \dfrac{ {( {2}^{a} \times {5}^{a} )}^{4} }{ {10}^{2} } = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

$\rm :\longmapsto\: \dfrac{ {(xy)}^{4} }{ {10}^{2} } = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

We know,

$\red{\boxed{ \rm{ {(xy)}^{a} = {x}^{a} \times {y}^{a}}}}$

So, using this

$\rm :\longmapsto\: \dfrac{ {(x)}^{4} {(y)}^{4} }{ {10}^{2} } = \dfrac{ {x}^{ \alpha } {y}^{ \beta } }{ {10}^{ \gamma } }$

On comparing both sides, we get

$\rm :\longmapsto\: \alpha = 4$

$\rm :\longmapsto\: \beta = 4$

$\rm :\longmapsto\: \gamma = 2$

Now, Consider,

$\rm :\longmapsto\:\dfrac{ \alpha + \beta }{ \gamma }$

$\rm \: = \: \dfrac{4 + 4}{2}$

$\rm \: = \: \dfrac{8}{2}$

$\rm \: = \: 4$

Hence,

$\bf:\longmapsto\:\dfrac{ \alpha + \beta }{ \gamma } = 4$

Additional Information :-

$\red{\boxed{ \rm{ log(xy) = logx + logy}}}$

$\red{\boxed{ \rm{ log( \frac{x}{y} ) = logx - logy}}}$

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

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

$\red{\boxed{ \rm{ log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y}}}}$

$\red{\boxed{ \rm{ {e}^{logx} = x}}}$

$\red{\boxed{ \rm{ {e}^{ylogx} = {x}^{y} }}}$