What is the answer to 2^r=(1/3)^{r-1} ?

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

Solve for r: 2^r=(1/3)^{r-1}

The solution is r=(ln(3))/(ln(6))

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2^{r} = (\frac{1}{3})^{r - 1}

r = \frac{ln ( 3 )}{ln ( 6 )}

Solution steps

2^{r} = (\frac{1}{3})^{r - 1}

Apply exponent rules

r ln (2) = (r - 1) ln (\frac{1}{3})

Solve

r ln (2) = (r - 1) ln (\frac{1}{3}) : r = \frac{ln ( 3 )}{ln ( 6 )}

r = \frac{ln ( 3 )}{ln ( 6 )}

10 = 2^{\frac{t}{3}}

2 = 1. 2^{x}

5^{5 x + 2} = 15

e^{4 - 4 x} = 3

e^{x} = \frac{5}{2}

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