What is the answer to (8/7)^x=5^{1-x} ?

The answer to (8/7)^x=5^{1-x} is x=(ln(5))/(ln(40/7))

Solve for x: (8/7)^x=5^{1-x}

The solution is x=(ln(5))/(ln(40/7))

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(\frac{8}{7})^{x} = 5^{1 - x}

x = \frac{ln ( 5 )}{ln ( \frac{40}{7} )}

Decimal Notation

x = 0.92338 \dots

Solution steps

(\frac{8}{7})^{x} = 5^{1 - x}

Apply exponent rules

x ln (\frac{8}{7}) = (1 - x) ln (5)

Solve

x ln (\frac{8}{7}) = (1 - x) ln (5) : x = \frac{ln ( 5 )}{ln ( \frac{40}{7} )}

x = \frac{ln ( 5 )}{ln ( \frac{40}{7} )}

6 e^{1 - x} = 24 e^{x}

e^{x} = - \frac{5}{8}

x, e^{y x} = 0

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

2 5^{x + 2} = 5

What is the answer to (8/7)^x=5^{1-x} ?
The answer to (8/7)^x=5^{1-x} is x=(ln(5))/(ln(40/7))
Solve for x: (8/7)^x=5^{1-x}
The solution is x=(ln(5))/(ln(40/7))