What is 2log_{10}(x+1)-log_{10}(4x) ?

The solution to 2log_{10}(x+1)-log_{10}(4x) is log_{10}(((x+1)^2)/(4x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 2log_{10}(x+1)-log_{10}(4x)

simplify

2 lo g_{10} (x + 1) - lo g_{10} (4 x)

lo g_{10} (\frac{( x + 1 ) ^{2}}{4 x})

Solution steps

2 lo g_{10} (x + 1) - lo g_{10} (4 x)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

= lo g_{10} ((x + 1)^{2}) - lo g_{10} (4 x)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{10} ((x + 1)^{2}) - lo g_{10} (4 x) = lo g_{10} (\frac{( x + 1 ) ^{2}}{4 x})

= lo g_{10} (\frac{( x + 1 ) ^{2}}{4 x})

ln (40) - 2 ln (5)

e^{- 2 l n (l n (x)) - 1 l n (x) + c}

- lo g_{10} (2.51 \cdot 1 0^{- 7})

2 lo g_{10} (x - 3) - lo g_{10} (x + 1)

ln (x 50 0^{- x})

What is 2log_{10}(x+1)-log_{10}(4x) ?
The solution to 2log_{10}(x+1)-log_{10}(4x) is log_{10}(((x+1)^2)/(4x))