Question

let us show that
a^log x base a =x​

Answer 1

Step-by-step explanation:

apply log base a on both sides

logx base a * log a base a = log x base a

log a base a = 1

log x base a = log x base a

hence proved

hope it helps u

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Answer 2

Answer:

The value of  $a^{log_{a}x }$ = x.

Step-by-step explanation:

Given that $a^{log_{a}x }$

We have to show that  $a^{log_{a}x }$ = x.

Let y =  $a^{log_{a}x }$

Take log on both sides with base 'a'

log(y) = log( $a^{log_{a}x }$)

That is

$log_{a}y= log_{a}a^{log_{a}x } } \\log_{a}y= (log_{a} x)(log_{a}a)$

We know that $log_{e}e = 1$

$log_{a} y= log_{a} x$

Base same then powers are equal.

That is  y = x

Here y = $a^{log_{a}x }$

∴ $a^{log_{a}x }$ = x.

Hence proved.