What is (3log_{7}(m)+12log_{7}(n))-3log_{7}(p) ?

The solution to (3log_{7}(m)+12log_{7}(n))-3log_{7}(p) is log_{7}((m^3n^{12})/(p^3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (3log_{7}(m)+12log_{7}(n))-3log_{7}(p)

simplify

(3 lo g_{7} (m) + 12 lo g_{7} (n)) - 3 lo g_{7} (p)

lo g_{7} (\frac{m ^{3} n ^{12}}{p ^{3}})

Solution steps

(3 lo g_{7} (m) + 12 lo g_{7} (n)) - 3 lo g_{7} (p)

Apply log rule:

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

3 lo g_{7} (m) = lo g_{7} (m^{3})

= (12 lo g_{7} (n) + lo g_{7} (m^{3})) - 3 lo g_{7} (p)

Apply log rule:

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

12 lo g_{7} (n) = lo g_{7} (n^{12})

= (lo g_{7} (m^{3}) + lo g_{7} (n^{12})) - 3 lo g_{7} (p)

Apply log rule:

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

lo g_{7} (m^{3}) + lo g_{7} (n^{12}) = lo g_{7} (m^{3} n^{12})

= lo g_{7} (m^{3} n^{12}) - 3 lo g_{7} (p)

Apply log rule:

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

3 lo g_{7} (p) = lo g_{7} (p^{3})

= lo g_{7} (m^{3} n^{12}) - lo g_{7} (p^{3})

Apply log rule:

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

lo g_{7} (m^{3} n^{12}) - lo g_{7} (p^{3}) = lo g_{7} (\frac{m ^{3} n ^{12}}{p ^{3}})

= lo g_{7} (\frac{m ^{3} n ^{12}}{p ^{3}})

lo g_{6} (c a \cdot b)

- ln (\frac{4}{5})

6 lo g_{9} (z) + 6 lo g_{9} (x) - 2 lo g_{9} (y)

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

ln (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1})

What is (3log_{7}(m)+12log_{7}(n))-3log_{7}(p) ?
The solution to (3log_{7}(m)+12log_{7}(n))-3log_{7}(p) is log_{7}((m^3n^{12})/(p^3))