how can this be solved
2006-11-12 09:23:00 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
27 and 9 can both be written in base 3
27 = 3^3
9 = 3^2
So 27^(4x) = (3^3)^(4x) = 3^(12x)
9^(x+1) = (3^2)^(x+1) = 3^(2x+2)
When the bases are the same, then the exponents are equal.
Therefore,
12x = 2x + 2
10x = 2
x = 2/10
x = 1/5
2006-11-12 09:25:32 · answer #1 · answered by MsMath 7 · 5⤊ 4⤋
27 = 3^3
so 27^(4x) = (3^3)^(4x) = 3^(12x)
9 = 3^2
so 9^(x+1) = (3^2)^(x+1) = 3^(2x+2)
this gives us 3^(12x) = 3^(2x+2)
so 12x = 2x + 2
x = 1/5
2006-11-12 09:30:48 · answer #2 · answered by Demiurge42 7 · 5⤊ 0⤋
use a calculator for this problem
27 ^(4x) = 9 ^(x + 1)
multiply by log
log [27 ^(4x)] = log [9 ^(x + 1)]
x (log 27^4) = (x + 1)(log 9)
xlog 531441 = xlog 9 + log 9
xlog 531441 - xlog 9 = log 9
x(log 531441 - log 9) = log 9
x = log 9/(log 531441 - log 9)
use your calculator
x = 0.2
ANSWER
x = 0.2
2006-11-12 09:36:57 · answer #3 · answered by trackstarr59 3 · 5⤊ 0⤋
MsMath is correct! I also get the same answer as her.
2006-11-12 09:31:34 · answer #4 · answered by Anonymous · 5⤊ 0⤋