Solve log_{7}49text{(ii)}log_{6}1text{(iii)}log_{10}100

$$\log _ { 7 } 49 \text { (ii) } \log _ { 6 } 1 \text { (iii) } \log _ { 10 } 100$$

Evaluate

$$0$$

Solution Steps

The base 7 logarithm of 49 is 2.

$$2ii\log_{6}\left(1\right)iii\log_{10}\left(100\right)$$

Multiply 2i and i to get -2.

$$-2\log_{6}\left(1\right)iii\log_{10}\left(100\right)$$

The base 6 logarithm of 1 is 0.

$$-2\times 0ii\log_{10}\left(100\right)$$

Multiply -2 and 0 to get 0.

$$0i\log_{10}\left(100\right)$$

The base 10 logarithm of 100 is 2.

$$0\times 2$$

Multiply 0 and 2 to get 0.

$$0$$

Real Part

$$0$$

Solution Steps

The base 7 logarithm of 49 is 2.

$$Re(2ii\log_{6}\left(1\right)iii\log_{10}\left(100\right))$$

Multiply 2i and i to get -2.

$$Re(-2\log_{6}\left(1\right)iii\log_{10}\left(100\right))$$

The base 6 logarithm of 1 is 0.

$$Re(-2\times 0ii\log_{10}\left(100\right))$$

Multiply -2 and 0 to get 0.

$$Re(0i\log_{10}\left(100\right))$$

The base 10 logarithm of 100 is 2.

$$Re(0\times 2)$$

Multiply 0 and 2 to get 0.

$$Re(0)$$

The real part of 0 is 0.

$$0$$