Please explain!
2007-04-10 04:56:45 · 6 answers · asked by Arch-RF d 1 in Science & Mathematics ➔ Mathematics
wait is it (3^2x) +1 = (5^3x)-1??
or
3^(2x+1) = 5^ (3x-1)
ok.. people are giving you different answers because you didn't put parenthesis .~_~
2007-04-10 05:00:30 · answer #1 · answered by Jami 3 · 0⤊ 1⤋
I am assuming you meant: 3^(2x+1)=5^(3x-1).
Take the log of each side:
log(3^(2x+1)) = log(5^(3x-1))
Properties of logarithms state that when you take the log of something, exponents move down in front as multipliers:
(2x+1)log3 = (3x-1)log5
Distribute the log3 and log 5:
(2x)log3+log3 = (3x)log5-log5
Collect all terms with an "x" on the left side and all terms without an "x" on the right side:
(2x)log3-(3x)log5 = -log5-log3
Pull the common "x" factor out of each term on the left:
x(2log3-3log5) = -log5-log3
Divide out everything non-x on the left side to isolate x:
x = (log5-log3) / (2log3-3log5)
This is the exact answer. Putting the above into a calculator (be careful with parentheses!) will produce a decimal approximation.
2007-04-10 05:44:04 · answer #2 · answered by Anonymous · 1⤊ 0⤋
3^2x+1 = 5^3x-1 Solve for x.?
9x+1= 125x-1
9x+2=125x
2=116x
x=1/80
2007-04-10 05:03:17 · answer #3 · answered by Anonymous · 0⤊ 1⤋
1/16 = x
2007-04-10 05:02:13 · answer #4 · answered by Joey G 2 · 0⤊ 2⤋
Restate with parantheses in correct places or else no one is really sure of what you mean and we are all guessing.
2007-04-10 06:29:30 · answer #5 · answered by TBU 2 · 2⤊ 0⤋
9x+1=125x-1
116x=2
x=1/58
2007-04-10 05:01:23 · answer #6 · answered by iyiogrenci 6 · 0⤊ 3⤋