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

The answer to 5^{x+1}=4^{2x-1} is x=(ln(20))/(4ln(2)-ln(5))

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

The solution is x=(ln(20))/(4ln(2)-ln(5))

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

x = \frac{ln ( 20 )}{4 ln ( 2 ) - ln ( 5 )}

Decimal Notation

x = 2.57553 \dots

Solution steps

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

Apply exponent rules

(x + 1) ln (5) = (2 x - 1) ln (4)

Solve

(x + 1) ln (5) = (2 x - 1) ln (4) : x = \frac{ln ( 20 )}{4 ln ( 2 ) - ln ( 5 )}

x = \frac{ln ( 20 )}{4 ln ( 2 ) - ln ( 5 )}

2^{- x} = 4 \cdot 8^{- x}

e^{7 x - 1} = 5

(0.97)^{x} = 0.5

3^{x + 4} = 81

x, e^{- 2 x} - e^{2 y} = c

What is the answer to 5^{x+1}=4^{2x-1} ?
The answer to 5^{x+1}=4^{2x-1} is x=(ln(20))/(4ln(2)-ln(5))
Solve for x: 5^{x+1}=4^{2x-1}
The solution is x=(ln(20))/(4ln(2)-ln(5))