Solve 6log_{2}2-log_{2}2^6-log_{2}2^-3

$$6 \log _ { 2 } 2 - \log _ { 2 } 2 ^ { 6 } - \log _ { 2 } 2 ^ { - 3 }$$

Evaluate

$$3$$

Solution Steps

The base 2 logarithm of 2 is 1.

$$6\times 1-\log_{2}\left(2^{6}\right)-\log_{2}\left(2^{-3}\right)$$

Multiply 6 and 1 to get 6.

$$6-\log_{2}\left(2^{6}\right)-\log_{2}\left(2^{-3}\right)$$

Calculate 2 to the power of 6 and get 64.

$$6-\log_{2}\left(64\right)-\log_{2}\left(2^{-3}\right)$$

The base 2 logarithm of 64 is 6.

$$6-6-\log_{2}\left(2^{-3}\right)$$

Subtract 6 from 6 to get 0.

$$0-\log_{2}\left(2^{-3}\right)$$

Calculate 2 to the power of -3 and get \frac{1}{8}.

$$0-\log_{2}\left(\frac{1}{8}\right)$$

The base 2 logarithm of \frac{1}{8} is -3.

$$0-\left(-3\right)$$

The opposite of -3 is 3.

$$0+3$$

Add 0 and 3 to get 3.

$$3$$

Factor

$$3$$