What is the answer to 3^{x^2-4x}=4^{x^2-4x} ?

The answer to 3^{x^2-4x}=4^{x^2-4x} is x=0,x=4

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3^{x^2-4x}=4^{x^2-4x}

3^{x^{2} - 4 x} = 4^{x^{2} - 4 x}

x = 0, x = 4

Solution steps

3^{x^{2} - 4 x} = 4^{x^{2} - 4 x}

Apply exponent rules

(x^{2} - 4 x) ln (3) = (x^{2} - 4 x) ln (4)

Solve

(x^{2} - 4 x) ln (3) = (x^{2} - 4 x) ln (4) : x = 0, x = 4

x = 0, x = 4

3^{2 x} + 2 \times 3^{x} - 15 = 0

3^{5 x} = 34

2 e^{x} (x - 1) + 1 = 0

5^{5} = 312 5^{3 x + 1}

2^{2 x} - 5 \times 2^{x} + 4 = 0

What is the answer to 3^{x^2-4x}=4^{x^2-4x} ?
The answer to 3^{x^2-4x}=4^{x^2-4x} is x=0,x=4
Solve for x: 3^{x^2-4x}=4^{x^2-4x}
The solution is x=0,x=4