What is the answer to (3^{x+1})(4^{x+2})=5^{2x-1} ?

The answer to (3^{x+1})(4^{x+2})=5^{2x-1} is x=(ln(240))/(2ln(5)-ln(12))

Solve for x: (3^{x+1})(4^{x+2})=5^{2x-1}

The solution is x=(ln(240))/(2ln(5)-ln(12))

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(3^{x + 1}) (4^{x + 2}) = 5^{2 x - 1}

x = \frac{ln ( 240 )}{2 ln ( 5 ) - ln ( 12 )}

Decimal Notation

x = 7.46712 \dots

Solution steps

(3^{x + 1}) (4^{x + 2}) = 5^{2 x - 1}

Apply exponent rules

(x + 1) ln (12) + ln (4^{1}) = (2 x - 1) ln (5)

Solve

(x + 1) ln (12) + ln (4^{1}) = (2 x - 1) ln (5) : x = \frac{ln ( 240 )}{2 ln ( 5 ) - ln ( 12 )}

x = \frac{ln ( 240 )}{2 ln ( 5 ) - ln ( 12 )}

2 e^{x} = 7 e^{x} - 3

\frac{1}{64} = 2^{x}

12 5^{x - 1} = 2 5^{x}

x, x y e^{\frac{y}{x}} = 5

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

What is the answer to (3^{x+1})(4^{x+2})=5^{2x-1} ?
The answer to (3^{x+1})(4^{x+2})=5^{2x-1} is x=(ln(240))/(2ln(5)-ln(12))
Solve for x: (3^{x+1})(4^{x+2})=5^{2x-1}
The solution is x=(ln(240))/(2ln(5)-ln(12))