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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.
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.:p
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
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.
*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 :rolleyes2
-_- .. well.. i see helping ppl really pays off.. ive learned my lesson.