Arithmetic explanation

Im studying for the GRE and thesevarithmetic problems have me scratching my head

Can someone explain these to me? I feel like a complete retard

Why are they taking the sqt root of 3^8x from 3^4x wtf

This one too. No idea whats going on

If 5^sqrt(y) is 81.
then 1/(25^-sqrt(y)) = 5^(2*sqrt(y)) = 81^2= 6561
also if 3^8x = 3600 then 3^4x = 60, thus
(3^(x-1))^4 = 3^4x * 3^-4 = 60/81

thus the expression equals 60*81=4860

user this is easy ajd the solution you posted tells you everything you need to know; we have nothing new to offer you
how about instead of making a thread that's worse than hw threads, you just put in some effort and study?

Dude, i know, i just cant undetstand the arithmetic

then study arithmetic.

It clearly explains the problem dude. What's wrong with you?

because sqrt(x)=x^1/2
so 3^(4x)=3^(8x/2)=sqrt(3^(8x))

What the fuck, this is what the GRE is?

Math GRE

Not the general GRE math section I'm assuming.

desu if OP is going to take the math subject GRE and this is giving him trouble he might want to stop studying for the GRE and go back and brush up for a few months and then start specific GRE studying.

desu

WTF [math] desu [/math] is translated to desu...

SHIT
To
Be
Honest is translated to desu

even then, this is one of the easy problems
a few problems every exam are pretty difficult, those are what you should study for

try typing F A M

why?
senpai

>Dude, i know, i just cant undetstand the entire problem

Ahh, the GRE. Nothing more advanced than 10th grade Geometry but everything seems like a trick question. What's your major OP?

For the first problem review your Law of Exponents

(X^a)(X^b) = X^(a+b)
(X^a)/(X^b)=X^(a-b)
(X^(a))^b = X^(ab)

Remember those three rules. After that you want to look at the givens that they've given you and try to make all the numbers in your problem the same base as the ones in the givens. (Ex: not much we can do with 25 but we can make it 5^2). This helps us because the goal is to plug in the values we've been given.

There's not much more to it other than practicing. It might take some practice to look at 3^(4x) and realize that it's relevant because it's also (3^(8x))^1/2 which will then allow you to make your substitution.

Good luck friend!

Poster from above I missed this one.

For this problem be sure to "factor out" as much as you can. We know that we can pull out a 3^(32) out of both terms leaving us with (3)^4-1 = 80

We know that the only prime factor of 3^(32) will be three so we look at the 80. Just simply find the largest factor here. Helps if you use a factor tree (probably a billion youtube videos on the subject if you've forgotten) but you will see that 2 and 5 are the only prime factors in 80.

Don't sweat it brah, most of the general GRE math stuff you can master in a month or less.

Solve for x
Solve for y

Plug x and y into the 3rd equation to get your answer.

log both sides and use log properties to solve for x and y

Op here

Thank you user

I solved for x and y but I don't get the right answer.

Can someone show what they solved for x and y?

I got x = 0.93 and y = 7.46

Middle rule should just be substituted with 1/X^a=X^(-a).

your x and y are correct, you should get 4860.