Solve 3div7=321^r6

$$3 \div 7 = 321 ^ { r } 6$$

Solve for r

$$r=-\log_{321}\left(14\right)\approx -0.457261414$$

Steps Using Definition of Logarithm

Divide both sides by 6.

$$\frac{\frac{3}{7}}{6}=321^{r}$$

Express \frac{\frac{3}{7}}{6} as a single fraction.

$$\frac{3}{7\times 6}=321^{r}$$

Multiply 7 and 6 to get 42.

$$\frac{3}{42}=321^{r}$$

Reduce the fraction \frac{3}{42} to lowest terms by extracting and canceling out 3.

$$\frac{1}{14}=321^{r}$$

Swap sides so that all variable terms are on the left hand side.

$$321^{r}=\frac{1}{14}$$

Take the logarithm of both sides of the equation.

$$\log(321^{r})=\log(\frac{1}{14})$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$r\log(321)=\log(\frac{1}{14})$$

Divide both sides by \log(321).

$$r=\frac{\log(\frac{1}{14})}{\log(321)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$r=\log_{321}\left(\frac{1}{14}\right)$$