Help with a math question? It's logarithms and exponentials.?

Question

Please help me solve this equation. Having trouble with this.

3^2x-1=12^x-7

Thanks! =)

Answer 1

log 3^(2x - 1) = (2x -1) log 3
12 = 4 x 3, so log 12 = log 4 + log 3 and

log 12^(x - 7) = (x - 7) (log 4 + log 3)

So, when we equate 3^(2x - 1) = 12^(x - 7) and take their logs, we get this:

(2x - 1) log 3 = (x - 7) (log 4 + log 3).

Now we simply divide both sides by log 3 to obtain this:

(2x - 1) = (x - 7) [(log 4 + log 3) / log 3)]
(2x - 1) = (x - 7) [(log 4 / log 3) + 1]
(2x - 1) = (x - 7)(2.2618595)
(2x - 1) = 2.2618595 x - 15.8330165)
-1 + 15.8330165 = 0.2618595 x
14.8330165 = 0.2618595 x
14.8330165 / 0.2618595 = x
56.6449432 = x

To verify that this is correct, just substitute the calculated value in for x and see if the answer is reasonably close.

3^(2x -1) = 3^(112.2898864) ~ 3.766096913 x 10^53
12^(x - 7) = 12^(49.6449432) ~ 3.766098396 x 10^53

So, our answer, despite that fact that it produces huge numbers, seems to be correct.

Answer 2

24-6

Answer 3

take log on both sides.........

then (2x-1)log3=(x-7)log12

(2x-1)log3=(x-7)(log4+log3)

sub the values of log 3,log4 and solve.......