how would i solve
logx = 2log 3 + 3log 2 ???
2006-12-06 11:43:40 · 5 answers · asked by rachel123go 3 in Science & Mathematics ➔ Mathematics
log(x)=2log(3)+3log(2)
log(x)=log(3^2)+log(2^3)
log(x)=log(9)+log(8)
log(x)=log(72)
x=72
2006-12-06 11:46:43 · answer #1 · answered by Greg G 5 · 0⤊ 1⤋
logx = 2log 3 + 3log 2 = log(3^2) + log(2^3) = log9 + log8 =
log (9 x 8) = log 72
=>
x = 72
2006-12-06 19:48:19 · answer #2 · answered by Anonymous · 1⤊ 0⤋
log x = log 3^2 +log 2^3
log x = log 9 +log 8
log x = log (9 x 8)
log x = log (72)
logs get canceled.
So x = 72
2006-12-06 19:56:05 · answer #3 · answered by ? 3 · 0⤊ 0⤋
log(x) = 2log(3) + 3log(2)
log(x) = log(3^2) + log(2^3)
log(x) = log(9) + log(8)
log(x) = log(9 * 8)
log(x) = log(72)
x = 72
2006-12-06 20:06:59 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
To undo base 10 logarithms, raise 10 to the power of both sides, i.e. 10^logx=10^(2log3 + 3log2)
Then, 10^log cancels.
Therefore, x=10^(2log3+3log2)
2006-12-06 19:52:10 · answer #5 · answered by shadowsandfog 2 · 0⤊ 1⤋