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.!