Question

please send the solution to this question.​

Answer 1

Answer:

2^-loga

Step-by-step explanation:

here when you change last ste you get the same thing.

Answer 2

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

Solve,

$\rm :\longmapsto\: {5}^{logx} + 5 {x}^{log5} = 3 \: where \: the \: base \: of \: log \: is \: a$

$\: \: \: \: \: \: \rm \: a) \: x = {2}^{ - log(a) } \: where \: base \: of \: log \: is \: 6$

$\: \: \: \: \: \: \rm \: b) \: x = {2}^{ - log(a) } \: where \: base \: of \: log \: is \: 5$

$\: \: \: \: \: \: \rm \: c) \: x = {2}^{ - log(a) } \: where \: base \: of \: log \: is \: 4$

$\: \: \: \: \: \: \rm \: d) \: x = {2}^{ - log(a) } \: where \: base \: of \: log \: is \: 3$

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

Given equation is

$\rm :\longmapsto\: {5}^{logx} + 5 {x}^{log5} = 3 \: where \: the \: base \: of \: log \: is \: a$

can be rewritten as

$\rm :\longmapsto\: {5}^{ log_{a}(x) } + 5 {x}^{ log_{a}(5) } = 3$

We know,

$\boxed{ \bf{ \: {x}^{logy} = {y}^{logx}}}$

So, using this property, it can ve rewritten as

$\rm :\longmapsto\: {5}^{ log_{a}(x) } + 5 \times {5}^{ log_{a}(x) } = 3$

can be again rewritten as

$\rm :\longmapsto\:6 \times {5}^{ log_{a}(x) } = 3$

$\rm :\longmapsto\: {5}^{ log_{a}(x) } = \dfrac{3}{6}$

$\rm :\longmapsto\: {5}^{ log_{a}(x) } = \dfrac{1}{2}$

$\rm :\longmapsto\: {5}^{ log_{a}(x) } = {2}^{ - 1}$

We know,

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

Using this identity, on RHS, we get

$\rm :\longmapsto\: {5}^{ log_{a}(x) } = {5}^{ log_{5}( {2}^{ - 1} ) }$

$\bf\implies \: log_{a}(x) = log_{5}( {2}^{ - 1} )$

We know,

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

So, using this result, we get

$\bf\implies \:x = {a}^{ log_{5}( {2}^{ - 1} ) }$

We know,

$\boxed{ \bf{ \: {x}^{logy} = {y}^{logx}}}$

So, using this

$\bf\implies \:x = {2}^{ - 1log_{5}( a ) }$

$\bf\implies \:x = {2}^{ - log_{5}( a ) }$

$\bf\implies \:x = {2}^{ - log(a) } \: where \: base \: of \: log \: is \: 5$

See the proof of

$\boxed{ \bf{ \: {x}^{logy} = {y}^{logx}}}$

Consider,

$\rm :\longmapsto\: {x}^{logy}$

can be rewritten as

$\rm \: = \: \: {e}^{log {x}^{logy} }$

$\rm \: = \: \: {e}^{log y \: logx }$

$\rm \: = \: \: {e}^{log x\: logy }$

$\rm \: = \: \: {e}^{log {y}^{logx} }$

$\rm \: = \: \: {y}^{logx}$

Hence, Proved