Question

expand log ( m^3 n^4/ p^2 ) using the laws of lagorithm​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to expand the given logarithm.

$\rm = \log \bigg( \dfrac{ {m}^{3} {n}^{4} }{ {p}^{2} } \bigg)$

We know that:

$\rm: \longmapsto \log \bigg( \dfrac{a}{b} \bigg) = \log(a) - \log(b)$

Therefore, we get:

$\rm = \log ({m}^{3} {n}^{4} ) - log( {p}^{2} )$

We know that:

$\rm: \longmapsto \log(ab) = \log(a) + \log(b)$

Therefore:

$\rm = \log ({m}^{3}) + \log( {n}^{4} ) - log( {p}^{2} )$

$\rm =3 \log (m) + 4\log(n) - 2 \log(p)$

Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

$\rm 1. \: \: {a}^{n} = b \implies log_{a}(b) = n$

$\rm 2. \: \: log_{a}(1) = 0, \: a \neq0,1$

$\rm 3. \: \: log_{a}(a) = 1, \: a \neq0,1$

$\rm 4. \: \: log_{a}(x) = log_{a}(y) \implies x = y$

$\rm 5. \: \: log_{e}(x) = ln(x)$

$\rm6. \: \: log_{a}(x) + log_{a}(y) = log_{a}(xy)$

$\rm7. \: \: log_{a}(x) - log_{a}(y) = log_{a} \bigg( \dfrac{x}{y} \bigg)$

$\rm 8. \: \: log_{a}( {x}^{n} ) = n\log_{a}(x)$

$\rm 9. \: \: log_{a}(m) = \dfrac{ log_{b}(m) }{ log_{b}(a) },m > 0,b > 0,a \ne1,b \ne1$

$\rm 10. \: \: log_{a}(b) = \dfrac{1}{ log_{b}(a) }$