What is log_{10}(2x+1)-log_{10}(x) ?

The solution to log_{10}(2x+1)-log_{10}(x) is log_{10}((2x+1)/x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify log_{10}(2x+1)-log_{10}(x)

simplify

lo g_{10} (2 x + 1) - lo g_{10} (x)

lo g_{10} (\frac{2 x + 1}{x})

Solution steps

lo g_{10} (2 x + 1) - lo g_{10} (x)

Apply log rule:

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

lo g_{10} (2 x + 1) - lo g_{10} (x) = lo g_{10} (\frac{2 x + 1}{x})

= lo g_{10} (\frac{2 x + 1}{x})

- lo g_{10} (1.8 \cdot 1 0^{- 4})

\frac{5}{2} ln (\frac{15}{4})

lo g_{7} (343 x)

ln (\frac{6 x - 5}{x + 4})

What is log_{10}(2x+1)-log_{10}(x) ?
The solution to log_{10}(2x+1)-log_{10}(x) is log_{10}((2x+1)/x)