Practice test exponent problem

Hi,
(1/5)^m x (1/4)^18 = (1/5)^m x (1/2)^36 = 1/[2(10)^35]= 1/[2(2*5)^35] = (1/2)^36*(1/5)^35.
i.e. (1/5)^m x (1/2)^36 = (1/2)^36x(1/5)^35
So, m =35

Thank you kindly. Much appreciated.

devooght wrote:I am struggling with the following problem:

(1/5)^m x (1/4)^18 = 1/[2(10)^35] m=?

Any advise?

Frankenstein's solution is correct. Would just like to point out the general case: Whenever you have an equation with variables in the exponents, your goal is usually to rewrite the powers in a different, yet equivalent, way so that the bases are the same on both sides of the equation.

This is most commonly done by
a) "Moving" powers from the exponent to the base and vice versa (as in rewriting 3^4 as 9^2, or 81^1), and/ or by
b) Splitting compound bases to their prime factors under the same exponent (as in splitting 10^25 into (2*5)^35 = 2^35*5^35.

Note that in the question above, the 2s don't actually interest you at all - only the 5s have the exponent of m required by the question. Once you split 10^35 so that you have 5^m = 5^35, the question is solved - the powers of 2 are irrelevant.

Hey frankenstine, thanks for solving.
I didn't get how you got the answer!
Could you write only 1 = sign in one line!
Thanks!

winniethepooh wrote:Hey frankenstine, thanks for solving.
I didn't get how you got the answer!
Could you write only 1 = sign in one line!
Thanks!

Okay..

LHS : (1/5)^m x (1/4)^18 = (1/5)^m x (1/2)^36
RHS : 1/[2(10)^35]
= 1/[2(2*5)^35]
=1/[2*(2^35)*(5^35)]
= (1/2)^36*(1/5)^35.
Let's equate the simplified forms,
i.e. (1/5)^m x (1/2)^36 = (1/2)^36x(1/5)^35
So, m =35

(1/5)^m x (1/4)^18 = 1/[2(10)^35]

1/5 ^m x 1/4 ^18 = 1/[2 * 2^35 * 5^35]

1/5 ^m x 1/4 ^18 = 1/2^36 * 1/5^35

1/5 ^m x 1/4 ^18 = 1/4^18 * 1/5^35

m=35