What is log_{10}(5/36)-log_{10}(5/9) ?

The solution to log_{10}(5/36)-log_{10}(5/9) is -2log_{10}(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify log_{10}(5/36)-log_{10}(5/9)

simplify

lo g_{10} (\frac{5}{36}) - lo g_{10} (\frac{5}{9})

- 2 lo g_{10} (2)

Decimal Notation

- 0.60205 \dots

Solution steps

lo g_{10} (\frac{5}{36}) - lo g_{10} (\frac{5}{9})

Apply log rule:

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

lo g_{10} (\frac{5}{36}) - lo g_{10} (\frac{5}{9}) = lo g_{10} (\frac{\frac{5}{36}}{\frac{5}{9}})

= lo g_{10} (\frac{\frac{5}{36}}{\frac{5}{9}})

Cancel

\frac{\frac{5}{36}}{\frac{5}{9}} : \frac{1}{4}

= lo g_{10} (\frac{1}{4})

\frac{1}{4} = 4^{- 1}

= lo g_{10} (4^{- 1})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

lo g_{10} (4^{- 1}) = - 1 \cdot lo g_{10} (4)

= - 1 \cdot lo g_{10} (4)

Apply rule:

1 \cdot a = a

= - lo g_{10} (4)

Factor the number:

4 = 2^{2}

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

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

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

= - 2 lo g_{10} (2)

- \frac{1}{44.6} ln (\frac{1}{2})

lo g_{5} (\frac{x}{x + 2})

5 ln (\frac{0.7}{3})

- lo g_{10} (6.4 x 1 0^{- 5})

ln (\frac{81}{25})

What is log_{10}(5/36)-log_{10}(5/9) ?
The solution to log_{10}(5/36)-log_{10}(5/9) is -2log_{10}(2)