What is log_{9}(z)+1/2 log_{9}(x)+1/2 log_{9}(y) ?

The solution to log_{9}(z)+1/2 log_{9}(x)+1/2 log_{9}(y) is log_{9}(zx^{1/2}y^{1/2})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify log_{9}(z)+1/2 log_{9}(x)+1/2 log_{9}(y)

simplify

lo g_{9} (z) + \frac{1}{2} lo g_{9} (x) + \frac{1}{2} lo g_{9} (y)

lo g_{9} (z x^{\frac{1}{2}} y^{\frac{1}{2}})

Solution steps

lo g_{9} (z) + \frac{1}{2} lo g_{9} (x) + \frac{1}{2} lo g_{9} (y)

Apply log rule:

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

\frac{1}{2} lo g_{9} (x) = lo g_{9} (x^{\frac{1}{2}})

= lo g_{9} (z) + lo g_{9} (x^{\frac{1}{2}}) + \frac{1}{2} lo g_{9} (y)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{9} (z) + lo g_{9} (x^{\frac{1}{2}}) = lo g_{9} (z x^{\frac{1}{2}})

= lo g_{9} (z x^{\frac{1}{2}}) + \frac{1}{2} lo g_{9} (y)

Apply log rule:

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

\frac{1}{2} lo g_{9} (y) = lo g_{9} (y^{\frac{1}{2}})

= lo g_{9} (z x^{\frac{1}{2}}) + lo g_{9} (y^{\frac{1}{2}})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{9} (z x^{\frac{1}{2}}) + lo g_{9} (y^{\frac{1}{2}}) = lo g_{9} (z x^{\frac{1}{2}} y^{\frac{1}{2}})

= lo g_{9} (z x^{\frac{1}{2}} y^{\frac{1}{2}})

ln (6.5 \cdot 1 0^{- 17})

ln (\frac{( 1 + x )}{1 - x})

ln (3) + 7 ln (x) - 2 ln (y)

lo g_{5} (\frac{4 a b ^{3} c ^{4}}{d ^{2}})

lo g_{3} (c) + lo g_{3} (c + 6)

What is log_{9}(z)+1/2 log_{9}(x)+1/2 log_{9}(y) ?
The solution to log_{9}(z)+1/2 log_{9}(x)+1/2 log_{9}(y) is log_{9}(zx^{1/2}y^{1/2})