exponents - no clue

This topic has expert replies
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Tue Jul 03, 2007 12:37 pm
Location: switzerland

exponents - no clue

by gian-28 » Tue Jul 03, 2007 12:58 pm
does anyone know how to solve this problem? it's totally freaking me out...

(1/5)^m *(1/4)^18=1/2(10)^35

I don't even know how to approach this one. it's the first problem of a practice test.. :evil:
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 277
Joined: Sun Jun 17, 2007 2:51 pm
Location: New York, NY
Thanked: 6 times
Followed by:1 members

by givemeanid » Tue Jul 03, 2007 3:39 pm
(1/5)^m * (1/4)^18 = 1/2 * 1/(10)^35

(1/5)^m = 1/2 * (1/10)^35 / (1/4)^18
= (4)^18 * 1/ 2 * (1/10)^35
= (4/10)^18 * 1/2 * (1/10)^17
= (2/5)^18 * 1/2 * (1/10)^17
= (1/5)^18 * 2^18 * 1/2 * (1/10)^17
= (1/5)^18 * 2^17 * (1/10)^17
= (1/5)^18 * (2/10)^17
= (1/5)^18 * (1/5)^17
= (1/5)^35

m = 35

Junior | Next Rank: 30 Posts
Posts: 18
Joined: Thu Jun 21, 2007 4:27 am
Location: New York

by sykedaddy » Wed Jul 04, 2007 1:37 pm
Easiest way for me was to rewrite...I just dislike fractions...

(1/5)^m *(1/4)^18=1/2(10)^35

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

..........................= (2)^-36 * (5)^-35

take out both (2)^-36 and

5^-m = 5^-35
No WAMMYs!!

User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Mon Dec 10, 2007 1:35 pm
Location: London, UK
Thanked: 19 times
GMAT Score:680

by II » Sat Dec 29, 2007 8:31 am
givemeanid wrote:(1/5)^m * (1/4)^18 = 1/2 * 1/(10)^35

(1/5)^m = 1/2 * (1/10)^35 / (1/4)^18
= (4)^18 * 1/ 2 * (1/10)^35
= (4/10)^18 * 1/2 * (1/10)^17
= (2/5)^18 * 1/2 * (1/10)^17
= (1/5)^18 * 2^18 * 1/2 * (1/10)^17
= (1/5)^18 * 2^17 * (1/10)^17
= (1/5)^18 * (2/10)^17
= (1/5)^18 * (1/5)^17
= (1/5)^35

m = 35
Apologies for the basic question, but how did you get from the question:

(1/5)^m *(1/4)^18=1/2(10)^35
to
(1/5)^m * (1/4)^18 = 1/2 * 1/(10)^35

The right hand side of the equation starts with 1/2(10)^35
And you have it down as 1/2 * 1/(10)^35
why ?

Thanks in advance.
II