By definition of logarithm we get

$x={ 6 }^{ 2.5 }={ 6 }^{ 2+0.5 }={ 6 }^{ 2 }.{ 6 }^{ 0.5 }={ 6 }^{ 2 }.{ 6 }^{ \cfrac { 1 }{ 2 } }=36\sqrt { 6 }$

Hence the answer.

Did it vai programming

include <iostream>

include <cmath>

include<stdio.h>

using namespace std;

int main() {

double a,x,val;


cout<<"Write the base number "; cin>>a; cout<<endl;

cout<<"Write down the number whose log you want  "; cin>>x; cout<<endl<<endl;

val=(log(x))/(log(a)); cout<<" log of "<<x<<"  is  "<<val<<endl;

double val2;

cout<<"Now if you want to calculate x from value then write down 1st the value "; cin>>val;

x=pow(a,val);

cout<<endl<<endl<<" x = "<<x<<endl;

return 0;

}

X=6^2.5 6^2.5 =6^5/2 = 6^2+1/2 -> x= 6^2•6^1/2 :. X=36√6

X = 6^(2.5) = 6^(2) x 6^(1/2) = 36sqrt(6)

log $_{6}$ $x$ = $2.5$
$x$ = $6^{2.5}$
and $6^{2.5}$ = $6^{2}$ x $6^{0.5}$ = $36$ $\sqrt{6}$

by the algo 6^2.5=6^2 6^1/2=36 6^1/2

6 powered by 2.5, it just same like 6^2 x 6^0.5 = 36 V6 (V = root)

According to the relation,

x = 6^2.5 x = 6^(2+0.5) x = 6².6^(1/2) x = 36 × √6 x = 36√6

By logarithm definition,

log N to the base a <=> a^x=N By using this, we get...., 6^2.5= x => x= 36 * 6^1/2