Help with a math problem?

[cheesyPOOF5]

I have no idea how to do this. Can anyone help?

 

[OG LOCC NESS]

Just simplify the square roots to actual numbers and solve

 

[cheesyPOOF5]

Just simplify the square roots to actual numbers and solve

Although that works, I know that's not what the teacher wants.

 

[go185]

Plug it into a calculator?

 

[Sonicslasher]

divide by zero?

 

[cheesyPOOF5]

divide by zero?

 

[joemommasfat]

just like (2^3)^4 = 2^12

(2^sqrt(2))^sqrt(18) = 2^sqrt(36) = 2^6

in general (a^b)^c = a^(b*c)

 

[cheesyPOOF5]

just like (2^3)^4 = 2^12

(2^sqrt(2))^sqrt(18) = 2^sqrt(36) = 2^6

in general (a^b)^c = a^(b*c)

Wow, thanks. That makes a lot of sense.

 

[rhyguy]

answer is

Spoiler

64

 

[FAST6191]

joemommasfat's method is correct but I suggest you learn the "power laws" as that is what this seems to want to teach.

2^2*2^2=2^4 aka a^x.a^y = a^(x+y)
You might like to think of this for the original purpose it was designed for.
Mathematicians and everyone really is lazy so if something can be written in shorthand it will, not to mention a large group of letters can be very confusing at a glance.
so a.a is written as a^2
a.a.a is written as a^3
a.a.a.a is written as a^4

divided by you take instead of adding a^x/a^y = a^(x-y) (instead you might like to think of it by doing the old method of if you have a divide put it 1/number and multiply it instead and when you know 1/a = a^-1 is becomes a bit easier)

square numbers are inverse powers
i.e. square root of 2 is 2^1/2

power on power gets multiplied
(2^2)^2
=2^4

hope you get it sorted.