Solve 4-2log_{9}(1/3)+4e^ln(3)

$$4 - 2 \log _ { 9 } ( \frac { 1 } { 3 } ) + 4 e ^ { \ln ( 3 ) }$$

Evaluate

$$17$$

Solution Steps

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

$$4-2\left(-\frac{1}{2}\right)+4e^{\log_{e}\left(3\right)}$$

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

$$4-\left(-1\right)+4e^{\log_{e}\left(3\right)}$$

The opposite of -1 is 1.

$$4+1+4e^{\log_{e}\left(3\right)}$$

Add 4 and 1 to get 5.

$$5+4e^{\log_{e}\left(3\right)}$$

Apply rule a^{\log_{a}\left(b\right)}=b, where a=e and b=3.

$$5+4\times 3$$

Multiply 4 and 3 to get 12.

$$5+12$$

Add 5 and 12 to get 17.

$$17$$

Factor

$$17$$