If 8^2-x = 4^3x, x=?
2007-12-11 05:32:53 · 3 answers · asked by pinkduck000 2 in Science & Mathematics ➔ Mathematics
please show your work or explain it
2007-12-11 05:36:10 · update #1
thankyou =)
2007-12-11 05:45:17 · update #2
8^(2 - x) = 4^(3x)
(2 - x) log 8 = 3x log 4
(2 - x) / 3x = log 4 / log 8
(2 - x) / 3x = log 2² / log 2³
(2 - x) / 3x = 2 log 2 / 3 log 2
(2 - x) / 3x = 2/3
6 - 3x = 6x
9x = 6
x = 2/3
2007-12-11 06:08:15 · answer #1 · answered by Como 7 · 3⤊ 0⤋
I'm assuming that's 8^(2-x).
Take the log base 2 of both sides. Then you have
log_2(8^2-x) = log_2(4^3x)
2-x(log_2(8))=3xlog_2(4)
3(2-x)=2(3x)
6-3x=6x
x = 2/3
=)
2007-12-11 13:39:00 · answer #2 · answered by Anonymous · 0⤊ 0⤋
8=2^3, so 8^(2-x) = 2^3(2-x)
4^3x = 2^6x
3(2-x)=6x
6-3x=6x
6=9x
x = 2/3
2007-12-11 13:38:42 · answer #3 · answered by norman 7 · 0⤊ 0⤋