What is 2log_{10}(a)-log_{10}(b) ?

The solution to 2log_{10}(a)-log_{10}(b) is log_{10}((a^2)/b)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 2log_{10}(a)-log_{10}(b)

simplify

2 lo g_{10} (a) - lo g_{10} (b)

lo g_{10} (\frac{a ^{2}}{b})

Solution steps

2 lo g_{10} (a) - lo g_{10} (b)

Apply log rule:

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

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

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

Apply log rule:

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

lo g_{10} (a^{2}) - lo g_{10} (b) = lo g_{10} (\frac{a ^{2}}{b})

= lo g_{10} (\frac{a ^{2}}{b})

lo g_{10} (\frac{11 x ^{3}}{4})

ln (sin (θ)) - ln (\frac{sin ( θ )}{6})

lo g_{10} (x) - 3 lo g_{10} (y) - 2 lo g_{10} (z)

\frac{1}{3} lo g_{10} (x) + 2 lo g_{10} (y)

lo g_{6} (5 z)

What is 2log_{10}(a)-log_{10}(b) ?
The solution to 2log_{10}(a)-log_{10}(b) is log_{10}((a^2)/b)