log5(x+3)+log5 4=log5 36
2007-05-23 08:00:41 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
loga(u)+loga(v)=loga(uv)
4x+12=36
x=6
2007-05-23 08:07:12 · answer #1 · answered by dismuhghettoname 4 · 0⤊ 0⤋
When you add logs of common bases, it's the same as multiplying the numbers.
log5(x + 3) + log5(4) = log5[4(x + 3)]
log5[4(x + 3)] = log5(36). The bases are the same, so the numbers are the same.
4(x + 3) = 36. Dividing by 4,
x + 3 = 9. Adding -3,
x = 6.
2007-05-23 15:06:26 · answer #2 · answered by Louise 5 · 0⤊ 0⤋
log5(x+3)+log5 4=log5 36
To begin with, you can combine things by using some simple log rules:
log5(x+3) + log5 4 = log5 (4x+12)
( log (a) + log (b) = log (a*b) )
So, now we have
log5 (4x+12) = log5 36
So, to get rid of the log5s, we raise each side to a power of 5.
5^[ log5 (4x+12) ] = 5^[ log5 36 ]
Which reduces down to
4x+12=36
Subtract 12 from each side
4x=24
Divide each side by 4
x=24/4=6
2007-05-23 15:30:54 · answer #3 · answered by oomps62 2 · 0⤊ 0⤋
move all but x to the right side. make it ((log5 36 - log5 4)/log5) - 3 = x (then just type it in your calculator. At least is what I'd do)
2007-05-23 15:06:18 · answer #4 · answered by German 2 · 0⤊ 0⤋
log5 4(x+3) = log5 36
4(x+3) = 36
x+3 = 9
x=6
2007-05-23 15:05:04 · answer #5 · answered by hawkeye3772 4 · 0⤊ 1⤋
OK, we know that log(a)+log(b)=log(a*b) so:
log5(x+3)+log5(4) = log5(4x+12).
Therefore 4x+12 = 36 (combining above line and original)
4x = 24
x =6.
2007-05-23 15:14:12 · answer #6 · answered by mikeburns55 5 · 0⤊ 0⤋
log5(x+3) = log5(36) - log5(4)
(x+3) = 5^(log5(36)-log5(4))
x = 5^(log5(36)-log5(4)) - 3
2007-05-23 15:05:18 · answer #7 · answered by Blank 2 · 0⤊ 0⤋
log (4(x+3)) = log 36
4x+12 = 36
4x= 24
x=6
2007-05-23 15:07:51 · answer #8 · answered by ironduke8159 7 · 0⤊ 1⤋
You need to manipulate the equation so that you have x on one side of the equation, X=
p.s. (works with any equation.)
2007-05-23 15:11:24 · answer #9 · answered by Ray T. 2 · 0⤊ 0⤋
Use log a+ log b=log a*b
2007-05-23 15:10:38 · answer #10 · answered by bamboo 2 · 0⤊ 0⤋