Solve 3log_{9}({3})-1div3log_{9}({3})

$$3 \log_{ 9 }({ 3 }) -1 \div 3 \log_{ 9 }({ 3 })$$

Evaluate

$$\frac{4}{3}\approx 1.333333333$$

Solution Steps

The base 9 logarithm of 3 is \frac{1}{2}.

$$3\times \frac{1}{2}-\frac{1}{3}\log_{9}\left(3\right)$$

Multiply 3 and \frac{1}{2} to get \frac{3}{2}.

$$\frac{3}{2}-\frac{1}{3}\log_{9}\left(3\right)$$

The base 9 logarithm of 3 is \frac{1}{2}.

$$\frac{3}{2}-\frac{1}{3}\times \frac{1}{2}$$

Multiply \frac{1}{3} and \frac{1}{2} to get \frac{1}{6}.

$$\frac{3}{2}-\frac{1}{6}$$

Subtract \frac{1}{6} from \frac{3}{2} to get \frac{4}{3}.

$$\frac{4}{3}$$

Factor

$$\frac{2 ^ {2}}{3} = 1\frac{1}{3} = 1.3333333333333333$$