Solve these for me pls

Solve these for me pls

nah

42

x=1 is number 1

for the first one
2^(3x)=8^x
2^3=8
8^x=8^x
true

number 2 is also x=1

#2 is both 1 and 3, but none in between

Lmao you posted this on /sci/ too. Do your own homework fag.

Protip: maybe if you at least attempt it we will help you

Don't ever ask homework help on Cred Forums, you moron

for #3 x=25

You have to show your work faggot

hAve you not learned logarithms?

Go back to /sci/

Number 4 is x=1.5, but I don't know what the (5/4)^x^2 is supposed to mean.

2^3x = 512^1/3x
2^3x = 2^3x

Well If you're going to be like that go fuck yourself
I ain't helping.

For the last one do
t = 2^-x
then 3t + 4t^2 - 1 = 0
t1 = 0.25 t2 = -1
then
2^-x = 0.25
x = -2
2^-x = -1
x = no answer

number 5 is x=2

x=2 not x=-2

Yep x = 2 my bad

When were you born? Not calling you an underage or anything, just curious.

bump

This looks like algebra 2, so I'd say somewhere around 14 to 16.

>reported.

Work for first one:
512 = 2^9
so
512^(1/(3x)) = 2^(9/(3x)) = 2^(3/x)
So 2^(3x) = 2^(3/x)
so
3x = 3/x
x^2 = 1
x = plus or minus 1