A student just emailed our headquarters with a terrific question, so I thought I'd post the response here for anyone interested:
While taking the practice test prescribed by
gmat.com, I cannot figure out how to do the
following problem. Can somebody please walk me
through this problem ?
The problem is:
(1/5)^m * (1/4)^18 = (1) / 2*(10)^35
This is a classic GMAT "reorganization" problem, in which your job is to rearrange the algebra to put it in a more user-friendly form. A few concepts can be extremely helpful on this kind of problem:
1) Whenever you have 1/"an exponent", that's the same thing as making the exponent negative and moving it to the numerator: 1/x = x^(-1). Using that logic, you can avoid having to carry fractions all the way through this and simplify the amount of work that you need to do.
2) When dealing with exponents with different bases, you'll almost always want to reduce the bases to their prime factors so that you can put them in like terms.
3) Answer choices can help you quite a bit in cases like these, too - if you look to the answer choices, you'll often get a hint as to what needs to be done.
Here, we're looking at:
5^m * 4^18 = 2^1 * 10^35
If we break 4 and 10, two of our exponent bases, into their prime factors, we'll get:
5^m * (2^2)^18 = 2 * (2*5)^35
Now, if we apply exponent rules, we can remove the parentheses by:
-Taking (2^2)^18 and expressing it as 2^(2*18) or 2^36
-Taking (2*5)^35 and expressing it as 2^35 * 5^35
5^m * 2^36 = 2 * 2^35 * 5^35
We can combine the bases of 2 on the right by noting that 2^1 * 2^35 = 2^36, so we have:
5^m * 2^36 = 2^36 * 5^35
The 2^36 terms on either side will cancel (divide both sides by 2^36 to do so), leaving us with:
5^m = 5^35
Therefore, m = 35.
One time-saving shortcut on this one: There is no way that 2 to any exponent will ever equal a multiple of 5, so you may even look at this one ahead of time and note that the twos are somewhat irrelevant. If 5^m * "something without a multiple of 5" = "something" * 5^35, the only thing that m can be is 35 to get both sides to equal, so noting that you can save a few algebraic steps.