can someone show me how to do this problem?
2007-12-07 09:53:25 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The only rule of logs you need for this is:
log(a) + log(b) = log(ab)
Use this rule on both sides:
log_2( 5x ) = log_2( 7(x-4) )
Raise both sides upon the power 2 to cancel the log_2:
5x = 7(x - 4)
From here it is straightforward algebra; distribute the 7:
5x = 7x - 28
Subtract 5x from both sides:
0 = 2x - 28
Add 28 to both sides:
28 = 2x
Divide both sides by 2:
x = 14
2007-12-07 10:19:48 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
log_2 5 + log_2 x = log_2 7 +log_2 (x-4)
log_2 5x = log_2 7(x-4)
5x = 7(x-4)
5x = 7x-28
-2x = -28
x = -28/-2 = 14
2007-12-07 10:03:06 · answer #2 · answered by Kyle K 3 · 0⤊ 0⤋
log2[5] + log2[x]=log2[7] + log2[x-4]
log2[5x] = log2[7x-28]
2^log2[5x] = 2^log2[7x-28]
5x = 7x-28
28 = 2x
14 = x
2007-12-07 10:05:38 · answer #3 · answered by J D 5 · 0⤊ 0⤋