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

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

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

The solution is x=(-ln(2)+ln(3))/(ln(3))

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3^{1 - x} = 2

x = \frac{- ln ( 2 ) + ln ( 3 )}{ln ( 3 )}

Solution steps

3^{1 - x} = 2

Apply exponent rules

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

Solve

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

x = \frac{- ln ( 2 ) + ln ( 3 )}{ln ( 3 )}

2^{4 x} = 16

2^{x} - 3^{x} - 3^{x} = 0

4 (10)^{x} = 20

2 e^{2 x + 7} - 15 = 0

e^{3 x} = 4

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