need help with pre-calculus.?

Question

Solve for x.
1. 3^(6x-2) = 7^x
2. 9^4x = 3^(x+3)

The ^ means (to the power of)

Answer 1

Well, to answer the questions, you should follow one basic rule: the bases should always be the same.

For number two:

9^4x = 3^(x+3)

The bases 9 and 3 are not the same and since

9 = 3^2

we use 3^2 instead. That gives us:

3^2(4x) = 3^(x+3)
3^8x = 3^(x+3)

Now that the bases are the same, we look for x.

8x = x+3

we add -x to both sides of the equation and that gives,

8x - x = 3
7x = 3

we divide seven to both sides,

x = 3/7 (x is equal to 3 over 7)

Checking:

9^4(3/7) = 3^(3/7 + 3)
9^12/7 = 3^24/7
3^2(12/7) = 3^24/7
3^24/7 = 3^ 24/7

Therefore, x = 3/7

And now you have an answer for number two!

Hmm.. I can't find a number where 3 and 7 would meet (I mean would become the same), so I'd leave number one to you.. Hehe..

Again, one basic rule: THE BASES SHOULD BE THE SAME BEFORE SOLVING FOR X.

However, I'm not sure if this is how your teacher asks you to solve it. This is only an alternative to the answer given above, the use of logarithms. Since your subject is pre-cal, use pre-cal ways on solving the problems. Use what your teacher has taught you. Good luck! :)

Answer 2

You solve power expressions using logarithms. Take the log of both sides (6x-2)*log(3) = x*log(7); solve for x. Same with the second part.

Answer 3

Well lets think about this for a second...

*REBOOT*