What is the answer to 5^{x+3}=7^{x-1} ?

The answer to 5^{x+3}=7^{x-1} is x=(ln(875))/(ln(7)-ln(5))

Solve for x: 5^{x+3}=7^{x-1}

The solution is x=(ln(875))/(ln(7)-ln(5))

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5^{x + 3} = 7^{x - 1}

x = \frac{ln ( 875 )}{ln ( 7 ) - ln ( 5 )}

Decimal Notation

x = 20.13308 \dots

Solution steps

5^{x + 3} = 7^{x - 1}

Apply exponent rules

(x + 3) ln (5) = (x - 1) ln (7)

Solve

(x + 3) ln (5) = (x - 1) ln (7) : x = \frac{ln ( 875 )}{ln ( 7 ) - ln ( 5 )}

x = \frac{ln ( 875 )}{ln ( 7 ) - ln ( 5 )}

3^{x} = 25

3^{x} = 20

e^{6 - 4 x} = 5

e^{2 x - 1} = 3

0.2 5^{x - 4} = 0. 5^{- 2 x + 1}

What is the answer to 5^{x+3}=7^{x-1} ?
The answer to 5^{x+3}=7^{x-1} is x=(ln(875))/(ln(7)-ln(5))
Solve for x: 5^{x+3}=7^{x-1}
The solution is x=(ln(875))/(ln(7)-ln(5))