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