What is the answer to 9^{x^2-7x}=27^{x^2-5} ?

The answer to 9^{x^2-7x}=27^{x^2-5} is x=-15,x=1

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9^{x^2-7x}=27^{x^2-5}

9^{x^{2} - 7 x} = 2 7^{x^{2} - 5}

x = - 15, x = 1

Solution steps

9^{x^{2} - 7 x} = 2 7^{x^{2} - 5}

Apply exponent rules

2 (x^{2} - 7 x) = 3 (x^{2} - 5)

Solve

2 (x^{2} - 7 x) = 3 (x^{2} - 5) : x = - 15, x = 1

x = - 15, x = 1

2^{x - 1} = 3^{2 x}

3^{x + 1} = 3^{3}

4^{(7 x - 2)} = 1 7^{(x + 1)}

e^{x - 1} - x - 1 = 0

e^{3 x} - 3 e^{2 x} + 4 e^{x} - 4 = 0

What is the answer to 9^{x^2-7x}=27^{x^2-5} ?
The answer to 9^{x^2-7x}=27^{x^2-5} is x=-15,x=1
Solve for x: 9^{x^2-7x}=27^{x^2-5}
The solution is x=-15,x=1