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

The answer to 2^{x+3}=5^{x-1} is x=(ln(40))/(ln(5)-ln(2))

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

The solution is x=(ln(40))/(ln(5)-ln(2))

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2^{x + 3} = 5^{x - 1}

x = \frac{ln ( 40 )}{ln ( 5 ) - ln ( 2 )}

Decimal Notation

x = 4.02588 \dots

Solution steps

2^{x + 3} = 5^{x - 1}

Apply exponent rules

(x + 3) ln (2) = (x - 1) ln (5)

Solve

(x + 3) ln (2) = (x - 1) ln (5) : x = \frac{ln ( 40 )}{ln ( 5 ) - ln ( 2 )}

x = \frac{ln ( 40 )}{ln ( 5 ) - ln ( 2 )}

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

10 e^{8 x + 3} + 2 = 8

3^{x + 4} = 2 7^{2 x - 1}

b^{2^{2 x - 17}} = b^{128}

What is the answer to 2^{x+3}=5^{x-1} ?
The answer to 2^{x+3}=5^{x-1} is x=(ln(40))/(ln(5)-ln(2))
Solve for x: 2^{x+3}=5^{x-1}
The solution is x=(ln(40))/(ln(5)-ln(2))