What is the answer to 2^{x^2+7x-38}=64^{x-6} ?

The answer to 2^{x^2+7x-38}=64^{x-6} is x=1,x=-2

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2^{x^2+7x-38}=64^{x-6}

2^{x^{2} + 7 x - 38} = 6 4^{x - 6}

x = 1, x = - 2

Solution steps

2^{x^{2} + 7 x - 38} = 6 4^{x - 6}

Apply exponent rules

x^{2} + 7 x - 38 = 6 (x - 6)

Solve

x^{2} + 7 x - 38 = 6 (x - 6) : x = 1, x = - 2

x = 1, x = - 2

79 = 3 3^{x}

15 = 20 + 30 e^{- \frac{t}{20}}

9^{x} = 3^{x} + 12

2000 = 8^{x}

(- 35)^{x} = 1

What is the answer to 2^{x^2+7x-38}=64^{x-6} ?
The answer to 2^{x^2+7x-38}=64^{x-6} is x=1,x=-2
Solve for x: 2^{x^2+7x-38}=64^{x-6}
The solution is x=1,x=-2