Eyes on Final Fantasy Forums > Reference > Study Hall > Exponential Equations

[Moon Rabbits]

How would I solve:

2 = 6(3 ^ 4f - 2)

27(3 ^ 3x + 1) = 3


I've tried a million different methods, but I can't figure out how to get rid of the 6(~) and 27(~) on the outside of the equations without leaving 2/6 or 3/27 on the other side of the equation, which I can't turn into a negative exponent because the numerators are not one.

I googled for a solution but only came up with answers involving "logs" which I haven't learned to apply yet.

[qwertyxsora]

For the first one

2 = 6(3 ^ (4f - 2))

(1/3) = 3^(4f - 2)

3^-1 = 3^(4f - 2)

-1 = 4f - 2

f = (1/4)

Remember, fractions mean negative exponents.

[Aerith's Knight]

10log1000=3
10^3=1000

so 3^4f-2=1/3

3log(1/3)=4f-2
(3log1/3 + 2) / 4 = f

edit: mine is more usefull.. because you dont need the same number in both exponentials

27(3 ^ 3x + 1) = 3

3/27 = 3^ (3x+1)

3log(3/27) = 3x+1

[3log(3/27) - 1] / 3 = x

[qwertyxsora]

She said she didn't learn to use logs yet, so I showed her how to do the first one without logs, and I think she can do the second one now that she can see how to do it.

[Moon Rabbits]

*is of the male variety*

Thanks a lot for your answer qwertyxsora. I didn't even think of reducing the fractions, instead I went through as many complicated answers as I could

[Aerith's Knight]

-_- .. well.. i see helping ppl really pays off.. ive learned my lesson.