How would I solve this one?
2 log3 3 + log3 (x-2) =4
2007-06-05 15:35:28 · 6 answers · asked by LicensedToIll 1 in Science & Mathematics ➔ Mathematics
first, you can evaluate the term 2log3 (3).
log3 (3) = 1.
2 log3(3) = 2.
Substituting in the original problem,
2 + log3(x-3) = 4.
subtract 2 from both sides:
log3(x-3) = 2.
raise both sides to exponent of base 3:
[ I will use '^' to mean exponent. For example, 2^3 = 2x2x2= 8 ]
(x-2) = 3 ^ 2 = 3 x 3 = 9
x-2 = 9
x = 11.
2007-06-05 15:43:30 · answer #1 · answered by old c programmer 4 · 1⤊ 0⤋
2log3 3 + log 3 (x-2 ) = 4
log3 ( 6 + ( x-2) ) = 4
log3 = 4
6x-12
log3=
2007-06-05 22:57:26 · answer #2 · answered by john w 1 · 0⤊ 0⤋
2 log3 3 + log3 (x-2) = 4
log3 3^2 +log3 (x-2) = 4
log3 9(x-2) = 4
3^4 = 9(x-2)
81 = 9x -18
99 = 9x
11 = x
2007-06-06 14:12:21 · answer #3 · answered by cubsmtdew 2 · 0⤊ 0⤋
lhs = log3 3^2+log3 (x-2)
= log3 9*(x-2)
9*(x-2) = 3^4 = 81
x-2 = 9
x=11
2007-06-05 22:51:03 · answer #4 · answered by r t 1 · 0⤊ 0⤋
In the following assume "log" means log to base 3.
2 log 3 + log (x - 2) = 4
log 3² + log (x - 2) = 4
log [ 3² (x - 2) ] = 4
3² (x - 2) = 3^4
9x - 18 = 81
9x = 99
x = 11
2007-06-06 03:22:36 · answer #5 · answered by Como 7 · 0⤊ 0⤋
I'm assuming those 3s right next to the log are base numbers.
2 log[3] (3) + log[3] (x-2) = 4
log[3] (3^2) + log[3] (x-2) = 4
log[3] (9) + log[3] (x-2) = 4
log[3] (9*(x-2)) = 4
9*(x-2) = 3^4
x-2 = 81/9
x = 2 + 9
x = 11
2007-06-05 22:38:55 · answer #6 · answered by Anonymous · 3⤊ 1⤋