What is the answer to 5^{x-1}=2^{2x+1} ?

The answer to 5^{x-1}=2^{2x+1} is x=(ln(10))/(ln(5)-2ln(2))

Solve for x: 5^{x-1}=2^{2x+1}

The solution is x=(ln(10))/(ln(5)-2ln(2))

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

x = \frac{ln ( 10 )}{ln ( 5 ) - 2 ln ( 2 )}

Decimal Notation

x = 10.31885 \dots

Solution steps

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

Apply exponent rules

(x - 1) ln (5) = (2 x + 1) ln (2)

Solve

(x - 1) ln (5) = (2 x + 1) ln (2) : x = \frac{ln ( 10 )}{ln ( 5 ) - 2 ln ( 2 )}

x = \frac{ln ( 10 )}{ln ( 5 ) - 2 ln ( 2 )}

3^{- 5} \cdot 3^{3} = 3^{k}

11 \cdot 1 0^{x - 3} - 1 0^{2 x - 4} = 0.1

5^{2 x} 3 = 75

x e^{x} - 7 x e^{x} = 0

What is the answer to 5^{x-1}=2^{2x+1} ?
The answer to 5^{x-1}=2^{2x+1} is x=(ln(10))/(ln(5)-2ln(2))
Solve for x: 5^{x-1}=2^{2x+1}
The solution is x=(ln(10))/(ln(5)-2ln(2))