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

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

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

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

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

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

Decimal Notation

x = 0.75647 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

3 x^{2} e^{x} + x^{3} e^{x} = 0

133.9 = \frac{133.9}{1 + 2 e ^{- t}}

e^{x^{2} - 4 x} = \frac{1}{e ^{2 x}}

2 = (1 + \frac{0.065}{12})^{12 t}

1 0^{x} = 1500

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