What is the answer to (9^{x^2-x})/(3^{3x-2)}=1 ?

The answer to (9^{x^2-x})/(3^{3x-2)}=1 is x= 1/2 ,x=2

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(9^{x^2-x})/(3^{3x-2)}=1

\frac{9 ^{x^{2} - x}}{3 ^{3 x - 2}} = 1

x = \frac{1}{2}, x = 2

Decimal Notation

x = 0.5, x = 2

Solution steps

\frac{9 ^{x^{2} - x}}{3 ^{3 x - 2}} = 1

Apply exponent rules

(2 (x^{2} - x) - 3 x + 2) ln (3) = ln (1)

Solve

(2 (x^{2} - x) - 3 x + 2) ln (3) = ln (1) : x = \frac{1}{2}, x = 2

x = \frac{1}{2}, x = 2

Verify Solutions:

x = \frac{1}{2}

True

, x = 2

True

The solutions are

x = \frac{1}{2}, x = 2

4^{(1 - x)} = 3^{(2 x + 3)}

e^{e^{e^{x}}} = 15.1546741867000 \dots

e^{x - 3} + 1 = 4 x - 3

3 e^{t} = 5 + 8 e^{- t}

8 = 3^{x}

What is the answer to (9^{x^2-x})/(3^{3x-2)}=1 ?
The answer to (9^{x^2-x})/(3^{3x-2)}=1 is x= 1/2 ,x=2
Solve for x: (9^{x^2-x})/(3^{3x-2)}=1
The solution is x= 1/2 ,x=2