What is the answer to 2*7^{r-4}=7 ?

The answer to 2*7^{r-4}=7 is r=(ln(7/2))/(ln(7))+4

Solve for r: 2*7^{r-4}=7

The solution is r=(ln(7/2))/(ln(7))+4

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2 \cdot 7^{r - 4} = 7

r = \frac{ln ( \frac{7}{2} )}{ln ( 7 )} + 4

Decimal Notation

r = 4.64379 \dots

Solution steps

2 \cdot 7^{r - 4} = 7

Divide both sides by

2

7^{r - 4} = \frac{7}{2}

Apply exponent rules

(r - 4) ln (7) = ln (\frac{7}{2})

Solve

(r - 4) ln (7) = ln (\frac{7}{2}) : r = \frac{ln ( \frac{7}{2} )}{ln ( 7 )} + 4

r = \frac{ln ( \frac{7}{2} )}{ln ( 7 )} + 4

2^{2 + x} + 2^{2 - x} = 17

28 = (100 - 20) 0. 5^{x} + 20

x, 5^{y} = 2^{x} + 1

9 - 3 x = (\frac{1}{27})^{x + 3}

9^{x} - 43 \cdot 3^{x - 1} = 0

What is the answer to 2*7^{r-4}=7 ?
The answer to 2*7^{r-4}=7 is r=(ln(7/2))/(ln(7))+4
Solve for r: 2*7^{r-4}=7
The solution is r=(ln(7/2))/(ln(7))+4