does anyone know the step by step process to solve the problem
(^3√6*^4√6)^12

I'm a little confused on your method of entering the problem. As it stands you have exponents which are unconnected to any number.

Currently this is what it reads like

to the 3 times root six multiplied by to the 4 times root 6 all of that to the 12th power.

You are either missing numbers or have the whole thing backwards. If your question is

12^((3√6)*(4√6))

then the first step is to get rid of roots in the exponent. Since I am assuming they are multiplied by each other to do that all you have to do is multiply in a manner I call inside- outsides. You want to multiply what is inside brackets roots and other special signifiers before you multiply or do anything else.

So the thing that is inside the most are the two sixes. Well you can't really mutiply them by the number outside of them but lucky for you

√6 x √6=√36 = 6.

So assuming that the equation I wrote is what you wanted to express then this is what you have now

12^(3*4*6)=12 ^ 72. And I will let you do the rest.

In general when doing exponents it is best to figure out what is in the exponents first and then work your way towards the number

for instance if I gave you this

12 ^ 3x

where x=10 you begin by multiplying 3 times 10 then go to the exponent. As it gets more difficult, keep the inside/outside rule in mind and it will stay relatively easy for instance

(12x)^2(((x-7)/3) + (10x)^2 - 14x^3)

where x= 4
In this case start with the one inside the most x-7/3= 4-7/3 = -1
go to farther outside
(10x)^2=1600
then even further
-14x^3=-896 NOTE: on this one only the x is cubed
Then continue the problem

Is this what you mean?


\left (
\overset {\small3\ } {\quad sqrt {6}}
\overset {\small4\ } {\quad sqrt {6}}
\right ) ^{12}


That is: the cube root of 6 times the fourth root of 6, all raised to the 12th power?

You can start by rewriting the cube root and the fourth root as follows:


\left ( 6^{\small {1/3}} 6 ^ {\small {1/4}} \right ) ^ {12}


Then apply the following rules:


a^b a^c = a^ {b+c}\\
(a*b)^c = a^c b^c\\
(a^b)^c = a^{bc}\\


Can you take it from here?