Question

To expand the quantity \(\ln \left( {{s^4}\sqrt {t\sqrt u } } \right)\) using logarithmic properties.

Short answer

The quantity \(\ln \left( {{s^4}\sqrt {t\sqrt u } } \right)\) can be expressed as \(4\ln (s) + \frac{1}{2}\ln (t) + \frac{1}{4}\ln (u)\).

Step-by-step solution

Given data

The given logarithm function is \(\ln \left( {{s^4}\sqrt {t\sqrt u } } \right)\).

Concept of Law of logarithm

Laws of logarithm:

Quotient law:\({\log _a}\left( {\frac{x}{y}} \right) = {\log _a}x - {\log _a}y\)

Product law:\({\log _a}(xy) = {\log _a}x + {\log _a}y\)

Power law:\({\log _a}\left( {{x^r}} \right) = r{\log _a}x\), where\(r\)is any real number.

Calculation of the value of \(\ln \left( {{s^4}\sqrt {t\sqrt u } } \right)\)

Simplify the given quantity \(\ln \left( {{s^4}\sqrt {t\sqrt u } } \right)\) as shown below.

\(\begin{array}{c}\ln \left( {{s^4}\sqrt {t\sqrt u } } \right) = \ln \left( {{s^4}{{(t\sqrt u )}^{\frac{1}{2}}}} \right)\\\ln \left( {{s^4}\sqrt {t\sqrt u } } \right) = \ln \left( {{s^4}{{\left( {t{{(u)}^{\frac{1}{2}}}} \right)}^{\frac{1}{2}}}} \right)\\\ln \left( {{s^4}\sqrt {t\sqrt u } } \right) = \ln \left( {{s^4}{t^{\frac{1}{2}}}{{(u)}^{\frac{1}{4}}}} \right)\end{array}\)

Use the law of logarithm and expand \(\ln \left( {{s^4}{t^{\frac{1}{2}}}{{(u)}^{\frac{1}{4}}}} \right)\) as follows:

\(\begin{array}{c}\ln \left( {{s^4}{t^{\frac{1}{2}}}{{(u)}^{\frac{1}{4}}}} \right) = \ln \left( {{s^4}} \right) + \ln \left( {{t^{\frac{1}{2}}}} \right) + \ln \left( {{u^{\frac{1}{4}}}} \right)\\\ln \left( {{s^4}{t^{\frac{1}{2}}}{{(u)}^{\frac{1}{4}}}} \right) = 4\ln (s) + \frac{1}{2}\ln (t) + \frac{1}{4}\ln (u)\end{array}\)

Thus, the quantity \(\ln \left( {{s^4}\sqrt {t\sqrt u } } \right)\) can be expressed as \(4\ln (s) + \frac{1}{2}\ln (t) + \frac{1}{4}\ln (u)\).