If (1/5)^m * (1/4)^18 = 1/(2*(10)^35), then m = ?

Thanks Bunnuel for the quick reply. So does that mean 1^m will always equal to 1? Because I was assuming m could be negative integer.

Yes, 1^m=1, for any m.

The negative exponent rule states: (1/a)^b can be re-expressed as a^(-b). Let’s flip all of the fractions using the negative exponent rule:

5^-m x 4^-18 = 2^-1 x 10^-35

5^-m x 2^-36 = 2^-1 x 2^-35 x 5^-35

5^-m x 2^-36 = 2^-36 x 5^-35

5^-m = 5^-35

m = 35

Answer: D

Exponent property #1: \((\frac{a}{b})^k=\frac{a^k}{b^k}\)

Exponent property #2: \((b^x)^y = b^{xy}\)

Exponent property #3: \((ab)^x = a^xb^x\)

------------------------------------------------------
Given: \((\frac{1}{5})^m * (\frac{1}{4})^{18} = \frac{1}{2*(10)^{35}}\)

Applying exponent property#1, we get: Given: \((\frac{1^m}{5^m})(\frac{1^{18}}{4^{18}}) = \frac{1}{2*(10)^{35}}\)

Simplify to get: Given: \((\frac{1}{5^m})(\frac{1}{4^{18}}) = \frac{1}{2*(10)^{35}}\)

ASIDE: To determine the value of m, we must rewrite both sides of the equation with similar base.
So, for the left side, we'll rewrite \(4\) as \(2^2\)
For the right side, we'll rewrite \(10\) as \(2 \times 5\)

We get: \((\frac{1}{5^m})(\frac{1}{(2^2)^{18}}) = \frac{1}{2*(2 \times 5)^{35}}\)

Applying exponent property#2, we get: \((\frac{1}{5^m})(\frac{1}{(2^{36}}) = \frac{1}{2*(2 \times 5)^{35}}\)

Applying exponent property#3, we get: \((\frac{1}{5^m})(\frac{1}{(2^{36}}) = \frac{1}{2*(2^{35})(5^{35})}\)

Since \(2*(2^{35} = 2^{36}\), we can write: \((\frac{1}{5^m})(\frac{1}{(2^{36}}) = \frac{1}{(2^{36})(5^{35})}\)

NOTE: Notice that I really didn't need to spend so much time working on the powers of 2, since the variable m was the exponent of 5.
Had I focused solely on the powers of 5, I could have answered the question MUCH faster.

That said, I wanted to show all of the exponent properties at work.

Cheers,
Brent

 
\(­5^{-m}*2^{-36} = ­2^{-1}*2^{-35}*5^{-35}­\)­­­

\(­5^{-m}*2^{-36} = ­2^{-36}*5^{-35}­\)­­­

Or, \(­5^{-m} = ­5^{-35}­\)­­­

Or, \(m = 35\), Answer must be (D)
 

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