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Question: Express the following logarithms in terms of \(\log a,\log b{\text{ and }}\log c\) . \(\log \left(...

Express the following logarithms in terms of loga,logb and logc\log a,\log b{\text{ and }}\log c .
log(a23×b32)\log \left( {\sqrt[3]{{{a^2}}} \times \sqrt[2]{{{b^3}}}} \right).

Explanation

Solution

Hint-In this question, we use the concept of properties of logarithm. We use property log(xy)=logx+logy and log(xy)=ylogx\log \left( {xy} \right) = \log x + \log y{\text{ and }}\log \left( {{x^y}} \right) = y\log x . We have to express the question in terms of loga,logb and logc\log a,\log b{\text{ and }}\log c so we use properties of logarithm mentioned in above.

Complete step-by-step answer:
Now, we have to solve problem log(a23×b32)\log \left( {\sqrt[3]{{{a^2}}} \times \sqrt[2]{{{b^3}}}} \right) in term of loga,logb and logc\log a,\log b{\text{ and }}\log c.
We know term a23 and b32\sqrt[3]{{{a^2}}}{\text{ and }}\sqrt[2]{{{b^3}}} are in product inside of logarithm. So, first we separate variables a and b by using property of logarithm,log(xy)=logx+logy\log \left( {xy} \right) = \log x + \log y
log(a23×b32) log(a23)+log(b32)  \Rightarrow \log \left( {\sqrt[3]{{{a^2}}} \times \sqrt[2]{{{b^3}}}} \right) \\\ \Rightarrow \log \left( {\sqrt[3]{{{a^2}}}} \right) + \log \left( {\sqrt[2]{{{b^3}}}} \right) \\\
We can write as,
log(a23)+log(b32)\Rightarrow \log \left( {{a^{\dfrac{2}{3}}}} \right) + \log \left( {{b^{\dfrac{3}{2}}}} \right)
Now, we use another property log(xy)=ylogx\log \left( {{x^y}} \right) = y\log x
23log(a)+32log(b)\Rightarrow \dfrac{2}{3}\log \left( a \right) + \dfrac{3}{2}\log \left( b \right)
Now, we can see questions expressed in terms of loga,logb and logc\log a,\log b{\text{ and }}\log c.
So, the answer to the above problem is 23log(a)+32log(b)\dfrac{2}{3}\log \left( a \right) + \dfrac{3}{2}\log \left( b \right) .
Note-In such types of problems we use some important points to solve questions in an easy way. First we separate variables a and b by using the property of logarithm log(xy)=logx+logy\log \left( {xy} \right) = \log x + \log y because we have to express questions in terms of loga and logb. Then we have to solve power and take it outside from the log by using another property log(xy)=ylogx\log \left( {{x^y}} \right) = y\log x . So, we will get the required answer.