Tricky Exponents (from GMAT Software Prep)

[Guest]

Sorry, I forgot to attribute the source of the earlier post.

Here is the Question again:

5^21 x 4^11 = 2 x 10^n. What is n?

Thanks!

[sarora]

5 ^ 21 * 4 ^ 11 = 2 * 10 ^ n

5^21 * (2^2)^11=2*(2*5)^n

5^21*2^22=2*2^n*5^n

5^21*2^22=2^(n+1)*5^n

Comparing exponents,

n=21

[StaceyKoprince]

Please make sure to post the full text of the question, including answer choices. Thanks!

[Eric]

Does anyone have an efficient method of attack for the problem below.

[RonPurewal]

Does anyone have an efficient method of attack for the problem below.
(picture)

First of all, be sure that you understand what direct and inverse proportions are. If you have an intuitive concept of these, then you'll notice that the addition of more chemical B will slow the reaction, which means that you are going to need more of chemical A to speed the reaction back up. That narrows the problem to the last 2 answers right away.

The best way to solve the problem, rather than just narrowing it, is to use smart numbers. Say that you started off with the concentrations of both chemicals at 1 unit. Then, according to the problem statement, the rate of reaction is constant times A^2 divided by B (go look up 'direct proportion' and 'inverse proportion' in an algebra textbook, or in our strategy guide, if you don't understand this). We can let the constant also be 1, so that
rate = A^2 divided by B
so the rate is 1 to start with.
Increasing the concentration of B means that the concentration of B is now 2. Therefore,
A^2 divided by 2 must equal 1
so
A^2 = 2
so
A = sqrt(2) which, as you should memorize, is about 1.4.
Therefore A has gone from 1 to 1.4, an increase of about 40 per cent.