What is log_{10}(0.401)-log_{10}(0.8) ?

The solution to log_{10}(0.401)-log_{10}(0.8) is log_{10}(401/8)-2

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simplify log_{10}(0.401)-log_{10}(0.8)

simplify

lo g_{10} (0.401) - lo g_{10} (0.8)

lo g_{10} (\frac{401}{8}) - 2

Decimal Notation

- 0.29994 \dots

Solution steps

lo g_{10} (0.401) - lo g_{10} (0.8)

lo g_{10} (0.401) = lo g_{10} (\frac{401}{1000})

lo g_{10} (0.8) = lo g_{10} (\frac{4}{5})

= lo g_{10} (\frac{401}{1000}) - lo g_{10} (\frac{4}{5})

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{10} (\frac{401}{1000}) - lo g_{10} (\frac{4}{5}) = lo g_{10} (\frac{\frac{401}{1000}}{\frac{4}{5}})

= lo g_{10} (\frac{\frac{401}{1000}}{\frac{4}{5}})

\frac{\frac{401}{1000}}{\frac{4}{5}} = \frac{401}{800}

= lo g_{10} (\frac{401}{800})

Apply log rule:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

lo g_{10} (\frac{401}{800}) = lo g_{10} (401) - lo g_{10} (800)

= lo g_{10} (401) - lo g_{10} (800)

lo g_{10} (800) = 2 + 3 lo g_{10} (2)

= lo g_{10} (401) - (2 + 3 lo g_{10} (2))

Apply the distributive law:

- (a + b) = - a - b

- (2 + 3 lo g_{10} (2)) = - 2 - 3 lo g_{10} (2)

= lo g_{10} (401) - 2 - 3 lo g_{10} (2)

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

- 3 lo g_{10} (2) = - lo g_{10} (2^{3})

= lo g_{10} (401) - 2 - lo g_{10} (2^{3})

lo g_{10} (401) - lo g_{10} (2^{3}) = lo g_{10} (\frac{401}{8})

= lo g_{10} (\frac{401}{8}) - 2

lo g_{4} (7 x y)

7 (lo g_{4} (3) + lo g_{4} (y)) - lo g_{4} (2)

lo g_{10} (\frac{x}{y ^{7}})

lo g_{10} (1.54 x) lo g_{10} (1.02)

lo g_{d} (u^{8} v^{3})

What is log_{10}(0.401)-log_{10}(0.8) ?
The solution to log_{10}(0.401)-log_{10}(0.8) is log_{10}(401/8)-2