+0

# Exponent review problem

+8
111
2

Here is a question I did when learning algebra thorugh the AoPS Intro to Algebra TEXTBOOK

Don't worry, this is from the textbook I'm self-learning and I've solved it.

This is also from an AMC12 so it's not a violation of copyright since I'm citing it

It's good for a exponent refresher.

Latex isn't working so here's the photo. This should be relatively easy for the moderators and probably some of the regular users too

I did it in 3 minutes, my parents in like 30 seconds LOL off-topic
May 14, 2020
edited by hugomimihu  May 14, 2020
edited by hugomimihu  May 14, 2020

#1
+4

5^(48/x)                             1

_________________ =        ____

5^(26/x) * 25^(17/x)               25^2

5^(48/x - 26/x)                     1

_____________ =          ______

(5^2)^(17/x)                   25^2

5^( 22/x)                       1

_______   =             ________

5^(34/x)                     (5^2)^2

1                             1

_______   =          _____

5^(12/x)                   5^4

Which implies that    12/x = 4     ⇒   x  = 3   May 14, 2020
#2
0

THis is how I did it(or similar, this is from a few days ago, doing it now)

So we first convert all 25^whatevers into 5^somethings

so

5^-4 = 5^48/x OVER 5^26/x * 5*34/x = 5 ^(48-26-34)x = 5^-12x

so -12x = -4, x = 3

essentially the same as what you did 