What is the answer to (1/6)^x=(4/3)^{x+1} ?

The answer to (1/6)^x=(4/3)^{x+1} is x=-(ln(4/3))/(3ln(2))

Solve for x: (1/6)^x=(4/3)^{x+1}

The solution is x=-(ln(4/3))/(3ln(2))

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(\frac{1}{6})^{x} = (\frac{4}{3})^{x + 1}

x = - \frac{ln ( \frac{4}{3} )}{3 ln ( 2 )}

Decimal Notation

x = - 0.13834 \dots

Solution steps

(\frac{1}{6})^{x} = (\frac{4}{3})^{x + 1}

Apply exponent rules

x ln (\frac{1}{6}) = (x + 1) ln (\frac{4}{3})

Solve

x ln (\frac{1}{6}) = (x + 1) ln (\frac{4}{3}) : x = - \frac{ln ( \frac{4}{3} )}{3 ln ( 2 )}

x = - \frac{ln ( \frac{4}{3} )}{3 ln ( 2 )}

x, 16 + 3^{2 x} = 3^{v} + 2

1 1^{n} = 161051

3^{x + 2} = 3^{x} + 1994

5^{2 n - 1} - 2 5^{n - 1} = 100

3^{x} - 3^{x - 2} = 8

What is the answer to (1/6)^x=(4/3)^{x+1} ?
The answer to (1/6)^x=(4/3)^{x+1} is x=-(ln(4/3))/(3ln(2))
Solve for x: (1/6)^x=(4/3)^{x+1}
The solution is x=-(ln(4/3))/(3ln(2))