Question

Please give the answer with explanation as soon as possible . Spammer will be reported.​

Answer 1

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

We have to evaluate the logarithm.

Given That:

$\rm = \log \bigg( \dfrac{ {a}^{3} {b}^{2} }{ {c}^{4} } \bigg) + \log \bigg( \dfrac{b {c}^{3} }{ {a}^{2} } \bigg) - \log \bigg( \dfrac{a}{c} \bigg)$

We know that:

$\rm: \longmapsto \log(x) + \log(y) = \log(xy)$

$\rm: \longmapsto \log(x) - \log(y) = \log \bigg( \dfrac{x}{y} \bigg)$

Therefore, we get:

$\rm = \log \bigg( \dfrac{ {a}^{3} {b}^{2} }{ {c}^{4} } \times \dfrac{b {c}^{3} }{ {a}^{2} } \bigg) - \log \bigg( \dfrac{a}{c} \bigg)$

$\rm = \log \bigg( \dfrac{ {a}^{3 - 2} {b}^{2 + 1} }{ {c}^{4 - 3} } \bigg) - \log \bigg( \dfrac{a}{c} \bigg)$

$\rm = \log \bigg( \dfrac{a{b}^{3} }{c} \bigg) - \log \bigg( \dfrac{a}{c} \bigg)$

$\rm = \log \bigg( \dfrac{a{b}^{3} }{c} \div \dfrac{a}{c} \bigg)$

$\rm = \log \bigg( \dfrac{a{b}^{3} }{c} \times \dfrac{c}{a} \bigg)$

$\rm = \log( {b}^{3})$

As we know that:

$\rm: \longmapsto \log( {x}^{n}) = n \log(x)$

Therefore, we get:

$\rm = 3\log(b)$

Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

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

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

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

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

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

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

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

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

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

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