solve without log ... (3^5x)(9^(x^2))=27?

Question

thanks! :(

Answer 1

(3^5x)(3^2)^(x²) = 27

(3^5x)(3^2x²) = 27

3^(5x + 2x²) = 3^3

3^(5x + 2x²) * 3^(-3) = 1

3^(2x² + 5x - 3) = 1

3^0 = 1
so
2x² + 5x - 3 = 0
(2x - 1)(x + 3) = 0

x = 1/2 or x = -3

Answer 2

Convert all the terms to the same base (3 in this case)
3^5x*3^2^(x^2) = 3^3
3^5x*3^2x^2 = 3^3
Get rid of the bases, since they're the same:

5x + 2x^2 = 3
Solve the quadratic: 2x^2+5x-3 =0
(2x-1)(x+3)
x = -3, 1/2.

So... any of those should work. :D

Answer 3

you have to change base so base 3 looks good
first one is good
second base is 9 which is 3^2 so 9^x^2 Would be 3^2x^2
and 27 is 3^3
so just look at the exponents and you have the quadratic
2x^2+5x-3=0
which factors to (2x-1)(x+3)
so x=1/2 , -3

Answer 4

3^5x(9^(x^2) = 27

3^5x(3^2)^x^2 = 3^3

3^(5x)(3)^2x^2 = 3^3

3^(5x+2x^2) = 3^3

5x + 2x^2 = 3

2x^2 + 5x - 3 = 0

2x^2 + 6x - x - 3 = 0

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

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

x = -3 or 1/2

Answer 5

Hello

(3^5x)(9^(x^2))=27

(3^5x)((3^2)^(x^2))=3^3?

(3^5x)(3^(2x^2))=3^3?

Since we have the same bases = we can drop these.

5x + 2x^2 = 3

2x^2 + 5x - 3 = (x+3)(2x-1)

x = 1/2
x = -3

Hope this helps

Answer 6

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

Answer 7

(3^5x) (3^2x^2) = 3^3

(5x) (2x^2) = 3
10x^3 = 3
continue . . .