What is the answer to \sqrt[4]{125^{3x+2}}=25^{x-1} ?

The answer to \sqrt[4]{125^{3x+2}}=25^{x-1} is x=-14

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\sqrt[4]{125^{3x+2}}=25^{x-1}

x = - 14

Solution steps

Apply exponent rules

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

Solve

\frac{3}{4} (3 x + 2) = 2 (x - 1) : x = - 14

x = - 14

- 3 e^{2 x} + 16 = 5

4^{2 x + 1} = 32

417 e^{x} = 3888

2^{2 x - 1} = 32

x, 6 e^{2 x} = 9 e^{4 x}

What is the answer to \sqrt[4]{125^{3x+2}}=25^{x-1} ?
The answer to \sqrt[4]{125^{3x+2}}=25^{x-1} is x=-14
Solve for x: \sqrt[4]{125^{3x+2}}=25^{x-1}
The solution is x=-14