Precalculus: Need Help with a Logarithm problem, read details?

Question

Alright.

3^(5X) * 9^(X^2) = 27

What steps do I take to solve?

Answer 1

Step 1 Convert all bases to the same number

In this case, 3, 9 and 27 are all powers of 3

9 = 3² and 27 = 3³

So 3^(5x) * 9^(x²) = 27
3^(5x) * 3^(2*x²) = 3³
ie 3^(2x² + 5x) = 3³

Step 2

Equate indices

So 2x² + 5x = 3

Step 3

Solve

So 2x² + 5x - 3 = 0

(x + 3)(2x - 1) = 0

x = ½, -3

Answer 2

Take the log and everything becomes clear. (It would be clearer still if we were at a blackboard, but we've got to make do with typing).

log( 3^(5x)* 9^(x^2) ) = log(27)
log(3^(5x)) + log( 9^(x^2) ) = log(27)
5x * log(3) + (x^2) * log(9) = log(27)

If you use log to the base 3, this becomes
5x + 2*x^2 = 3
and the quadratic resolves to
(x+3)*(2x-1) = 0
or x is -3 or .5

Answer 3

let nl be log base 3 I don't feel like writing ln_3 all the time

3^(5X) * 9^(X^2) = 27
5x*ln(3)+x^2*ln(9)=ln(27)
ln(9)x^2+5*ln(3)x-ln(27)=0
2x^2+5x-3=0
(2x-1)(x+3)
2x=1
x=1/2
x=-3

Answer 4

9=3^2
9^(x^2)= 3^(2(x^2))= 3(2x^2)

3^(5x) * 3^(2x^2)=27=3^3
3^(5x+2x^2)=3^3
5x+2x^2=3
2x^2+5x-3=0
(2x-1)(x+3)=0
x=-3 or 1/2

Answer 5

Basically, you need to apply those two rules:

log (a * b) = log (a) + log (b)

log (a^n) = n * log (a)

You should be able to figure the rest.