How do you simplify $35^(log_35x)$ ?

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Answer 1 · 2016-09-24T10:13:50

$35^(log_35x)=x$ .

By the Defn. of $log$ fun., we can immediately answer that,

$35^(log_(35)x)=x$ .

To see, how this is so, let us recall the Defn. of $log$ fun. :

$log_bx=m iff b^m=x$

Now, let us subst. the value of $m$ from the L.H.S of " $iff$ " to the

R.H.S., to get, the Desired Result : $b^(log_bx)=x$

Alternatively,

Suppose that, $35^(log_(35)x)=y$ .

Taking $log_35$ of both the sides, we have,

$log_35 (35^m)=log_35 y, ..........(1)$ , where, $m=log_35x$ .

But, $log_35(35^m)=mlog_35 35=m*1=m$ .

Therefore, by $(1)$ , we get,

$m=log_35y, i.e., log_35x=log_35y$ .

As, $log$ fun. is $1-1$ , we have, $x=y, i.e., 35^(log_35x)=x$ .

Enjoy Maths.!

How do you simplify 35^(log_35x)35^(log_35x)?