# Discussion on Exponentials and Logarithms

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*Posted on: 09:03:06 AM on July 6, 2015*

**Posted By:**

**Nikita Masand**

How should I solve, (2^log base 6 18) (3^log base 6 3)
Ans is 6

## Reponses To This

*Posted on: 12:25:44 AM on October 26, 2015*

(2^log base 6 18) (3^log base 6 3)
=((2^log base 6 (6)*(3)) (3^log base 6 3)
=(2^((log base 6 6)+ (log base 6 3))(3^log base 6 3)
[using log base x (ab) = log base x a + log base x b]
=(2^(1 + (log base 6 3))(3^log base 6 3)
=(2)*(2)^(log base 6 3)(3^log base 6 3)...................(1)
Now, we know that
f(x)^(log base a g(x))= g(x)^(log base a f(x))
Therefore, equation (1) can be written as,
=(2)*(3)^(log base 6 2)(3^log base 6 3)
=(2)*(3)^((log base 6 2) + (log base 6 3)) ....... [using (a^m)*(a^n)=a*(m+n)]
=(2)*(3)^((log base 6 (2)*(3))
[using log base x a + log base x b = log base x (ab)]
=(2)*(3)^((log base 6 6)
=(2)*(3)^(1)
=6 (Ans.)

*Posted on: 11:40:47 PM on April 10, 2017*

{2^(log base 6 18)} * {3^(log base 6 3)}
{18^(log base 6 2)} * {3^(log base 6 3)} -----since{a^(log base b c)}=={c^(log base b a)}
{(3*6)^(log base 6 2)} * {3^(log base 6 3)}
{(3^log base 6 2)*(6^log base 6 2)} * {3^(log base 6 3)}
{(3^log base 6 2)* 2} * {3^(log base 6 3)} -------since {a^log base a x==x}
2*{(3^log base 6 2) * 3^(log base 6 3)}
2*{3^(log base 6 2 + log base 6 3)}
2*{3^(log base 6 (2*3)}
2*{3^(log base 6 6)}
2*{3^1}
2*3
6