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

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

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

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

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

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

Decimal Notation

x = - 7.21256 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

3^{x^{2} - 1} = 134

x^{x} = 100

y, e^{3 x} + x^{2} \cdot y + e^{y} = 0

7^{x} + 7^{x + 1} - 7^{x + 2} + 7^{x - 1} = - 286

(\frac{1}{2})^{x} = 4

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