What is the answer to 3^{x^2-5}-9^{x-1}=0 ?

The answer to 3^{x^2-5}-9^{x-1}=0 is x=3,x=-1

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3^{x^2-5}-9^{x-1}=0

3^{x^{2} - 5} - 9^{x - 1} = 0

x = 3, x = - 1

Solution steps

3^{x^{2} - 5} - 9^{x - 1} = 0

Add

9^{x - 1}

to both sides

3^{x^{2} - 5} - 9^{x - 1} + 9^{x - 1} = 0 + 9^{x - 1}

Simplify

3^{x^{2} - 5} = 9^{x - 1}

Apply exponent rules

x^{2} - 5 = 2 (x - 1)

Solve

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

x = 3, x = - 1

5^{x - 2} = 4^{x + 1}

3^{(5 x - 3)} = 2 7^{(x + 1)}

e^{x} - e^{b - x} = c

5^{n} \cdot 5^{n + 4} = 25

2 7^{2 x + 3} = 9^{3}

What is the answer to 3^{x^2-5}-9^{x-1}=0 ?
The answer to 3^{x^2-5}-9^{x-1}=0 is x=3,x=-1
Solve for x: 3^{x^2-5}-9^{x-1}=0
The solution is x=3,x=-1