(5^21)*(4^11)

[guest]

Not sure how to simplify and solve for N

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

[mclaren7]

(5^21)*(4^11)=2*(10^n)
(5^21)*(2^2)^11=2*(5*2)^n
(5^21)*(2^22)=2*(5^n*2^n)
(5^21)*(2^22)=5^n*2^n+1

n= 21.

KH

[RonPurewal]

(5^21)*(4^11)=2*(10^n)
(5^21)*(2^2)^11=2*(5*2)^n
(5^21)*(2^22)=2*(5^n*2^n)
(5^21)*(2^22)=5^n*2^n+1

n= 21.

KH

perfect.

remember the general theme here: always try to get a COMMON BASE whenever you're dealing with exponentials on both sides.
if you're having difficulty discerning what that common base should be, then it's always a good idea to use primes (as is done in this problem: note that the first step is to break 4 down into 2^2 and to break 10 down into 5*2).

[themarkac]

What do you do if you can't see the substitution?

[RonPurewal]

What do you do if you can't see the substitution?

you should be comfortable enough with prime factorizations and exponents that this becomes a non-issue.

this sort of problem - in which primes are combined in different ways, obscuring the fundamental similarity of the two sides of an equation - is a mainstay of the gmat. you need to become proficient at it, no ifs, ands, or buts.

* learn to break down numbers into primes
* perfect your command of laws of exponents, so that you can manipulate the powers smoothly and accurately

the upside is that these sorts of problems are fairly easy to spot: basically anything featuring a bunch of different smallish positive integers raised to powers is probably one of them. therefore, if you skim through the o.g. and look for all the problems bearing a superficial similarity to this one, you should find plenty to practice on.