2006-12-17 11:49:29 · 7 answers · asked by celticevening 1 in Science & Mathematics ➔ Mathematics
Uh, actually blondie, its algebra II.
I'm assuming (x + 3) is the argument.
This can be rewritten as 2^5 = x + 3
32 = x + 3
29 = x
2006-12-17 11:51:37 · answer #1 · answered by laffytaffychick13 1 · 0⤊ 0⤋
log base 2 of (x+3)=5
<=> (x + 3) = 2^5
<=> x = 29
2006-12-17 17:51:36 · answer #2 · answered by James Chan 4 · 0⤊ 0⤋
Still something missing unless you mean log 2^(x+3)=5
(x+3) log 2=5
x+3=5/log 2
x=3+5/log 2=19.61
edit: Maybe you mean use logs to solve 2^(x+3)=5 in which case
(x+3)log 2=log 5
x+3=log 5 / log 2
x=3+log 5 / log 2=5.322
2006-12-17 11:52:36 · answer #3 · answered by yupchagee 7 · 0⤊ 1⤋
2^5=x+3
32=x+3
x=32-3
=29
so x=29
2006-12-17 11:55:11 · answer #4 · answered by raj 7 · 0⤊ 0⤋
x+3 = 2^5
x+3 = 32
x = 32-3
x=29
2006-12-17 11:58:36 · answer #5 · answered by Mr Bean 5 · 0⤊ 0⤋
your notation is confusing--what do you mean? log base 2 of (x+3)?
pls clarify
2006-12-17 11:53:58 · answer #6 · answered by vertigosr37 1 · 0⤊ 0⤋
HOLY SH!T ROLMFAO!
WTF IS THAT? Who the hell can know that except some college proffesor? Jeezuz Peezuz
2006-12-17 11:51:26 · answer #7 · answered by Anonymous · 0⤊ 0⤋