math logarithm question..

8^(x + 2) = 3[4^(2x)]
[2^3]^(x +2) = 3[2^4x]
2^(3x + 6) = 3[2^4x]
Let 2^x = y, so 2^3x = y^3 and 2^4x = y^4.
The equation becomes
64y^3 = 3y^4
y^3(3y - 64) = 0
y = 0 (rej.)
y = 64/3, that is
2^x = 64/3
x = log(64/3)/log2 = 1.329/0.301= 4.415.

You can try to use log first.

log[ 8^(x+2)] = log {3[4^(2x)]}

(x+2) log 8 = log 3 + (2x) log 4 (log ab = log a + log b)

(3x+6) log 2 - 4x log 2 = log 3

6 log 2 - x log2 =log 3
x=6 - 1og 3 / log 2

2008-10-15 16:38:26 補充：
Sorry, typo error

x=6 - log 3 / log 2