given 2^2x=3, find 8x?

Answer 1

If you mean

2^(2x) = 3

then

log(2^(2x)) = log(3)
2x log(2) = log(3)
x = log(3)/(2log(2))

8x = 8log(3)/(2log(2)) = 4log(3)/log(2)

Answer 2

If 2^(2x) = 3, then we change this to logarithmic form.
A reminder that
b^a = c in logarithmic form is log[base b](c) = a

log[base 2](3) = 2x

Since we want to find 8x, then

8x = 4(2x) = 4(log[base 2](3))

Answer 3

I'm assuming the left side is 2^(2x), not (2^2)x. In that case, take the log of both sides:

log(2^(2x)) = log(3)
2x * log(2) = log(3)
2x = log(3) / log(2)
x = log(3) / (2*log(2))

It doesn't matter what base you use for the log. You could also simplify this to (1/2) * (log[base 2](3))

Answer 4

Good question. It tests your ability to pay attention.

2^(2x) =3 therefore ln 2^(2x) = ln 3 or (2x) ln2 = ln 3

finally x = 1/2 ln 3/2 and so 8x = 4 ln 3/2

Answer 5

You could take the log of both sides, getting 2x log 2 = log 3,
then divide both sides by log 2, then multiply both sides by 4

Answer 6

2^2x=3
=>4x=3
=> x=3/4
Hence 8x=(3/4)X8=6

Answer 7

Gotta have parentheses!

Do you mean (2^2)x = 3 or do you mean 2^(2x) =3

HTH

Charles

Answer 8

seriously i think you r juz trying to make us type diff maths solution...