What is 2log_{a}(y)-log_{a}(z) ?

The solution to 2log_{a}(y)-log_{a}(z) is log_{a}((y^2)/z)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 2log_{a}(y)-log_{a}(z)

simplify

2 lo g_{a} (y) - lo g_{a} (z)

lo g_{a} (\frac{y ^{2}}{z})

Solution steps

2 lo g_{a} (y) - lo g_{a} (z)

Apply log rule:

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

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

= lo g_{a} (y^{2}) - lo g_{a} (z)

Apply log rule:

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

lo g_{a} (y^{2}) - lo g_{a} (z) = lo g_{a} (\frac{y ^{2}}{z})

= lo g_{a} (\frac{y ^{2}}{z})

2 ln (x + 4) - 8 ln (x)

- lo g_{10} (6.3 \cdot 1 0^{- 6})

ln (\frac{x ^{2}}{2})

\frac{ln ( 2 )}{5 ln ( 2 ) - ln ( 7 )}

\frac{ln ( x ^{3 - 1} ) - ln ( x ^{2} - x + 1 )}{( ln ( x + 1 ) )}

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