Solve log_{10}(10^-5/10)+log_{10}(10)-log_{10}(10^2)=

$$\log _ { 10 } ( \frac { 10 ^ { - 5 } } { 10 } ) + \log _ { 10 } ( 10 ) - \log _ { 10 } ( 10 ^ { 2 } ) =$$

Evaluate

$$-7$$

Solution Steps

Rewrite 10 as 10^{-5}\times 10^{6}. Cancel out 10^{-5} in both numerator and denominator.

$$\log_{10}\left(\frac{1}{10^{6}}\right)+\log_{10}\left(10\right)-\log_{10}\left(10^{2}\right)$$

Calculate 10 to the power of 6 and get 1000000.

$$\log_{10}\left(\frac{1}{1000000}\right)+\log_{10}\left(10\right)-\log_{10}\left(10^{2}\right)$$

The base 10 logarithm of \frac{1}{1000000} is -6.

$$-6+\log_{10}\left(10\right)-\log_{10}\left(10^{2}\right)$$

The base 10 logarithm of 10 is 1.

$$-6+1-\log_{10}\left(10^{2}\right)$$

Add -6 and 1 to get -5.

$$-5-\log_{10}\left(10^{2}\right)$$

Calculate 10 to the power of 2 and get 100.

$$-5-\log_{10}\left(100\right)$$

The base 10 logarithm of 100 is 2.

$$-5-2$$

Subtract 2 from -5 to get -7.

$$-7$$

Factor

$$-7$$