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

Express the following logarithms in terms of loga,logb,andlogc\log \,a,\,\log \,b,\,and\,\log \,c. log(a4b39)\log \left( \sqrt[9]{a{{-}^{4}}{{b}^{3}}} \right)

Explanation

Solution

Hint: Use logarithmic properties for solving this question logab=bloga;logab=loga+logb\log {{a}^{b}}=b\log a;\log ab=\log a+\log b.

Complete step-by-step solution -
Given expression in the question for which we need to simplify:
log(a4b39)\log \left( \sqrt[9]{{{a}^{-4}}{{b}^{3}}} \right)we can write the term inside the log as follow:
log((a4b3)19)\log \left( {{\left( {{a}^{-4}}{{b}^{3}} \right)}^{\dfrac{1}{9}}} \right)
By basic knowledge of logarithm, we can say that this is true:
logab=bloga\log {{a}^{b}}=b\log a
By applying the above formula, we get the equation as:
=19log(a4b3)=\dfrac{1}{9}\log \left( {{a}^{-4}}{{b}^{3}} \right)
By basic knowledge of logarithm, we can say that this is true:
logab=loga+logb\log ab=\log a+\log b
By applying the above formula, we get the equations:
=19(loga4+logb3)=\dfrac{1}{9}\left( \log {{a}^{-4}}+\log {{b}^{3}} \right)
Now divide the equation into 2 parts as 2 terms.
The 1st term be represented by A and value of A be given by
A=loga4A=\log {{a}^{-4}}
The 2nd term be represented B and the value of B is given by:
B=19logb3B=\dfrac{1}{9}\log {{b}^{3}}
Now simplification of first term (A) is given by as follows:
A=19log(a4)A=\dfrac{1}{9}\log \left( {{a}^{-4}} \right)
By basic knowledge of logarithm, we know the condition given by:
logab=bloga\log {{a}^{b}}=b\log a
By applying this here, we then the expression for A into:
A=49logaA=\dfrac{-4}{9}\log a
By keeping loga\log a in brackets we get final expression A as:
A=49(loga)A=\dfrac{-4}{9}\left( loga \right)
Now simplification of second term (B) b is given by as follow:
B=19log(b3)B=\dfrac{1}{9}\log \left( {{b}^{3}} \right)
By basic knowledge of logarithm, we know the condition as:
logab=bloga\log {{a}^{b}}=b\log a
By applying the above condition here, we turn B into:
B=39cosbB=\dfrac{3}{9}\cos b
By simplifying the above equation, we get final B as:
B=13logbB=\dfrac{1}{3}\log b
By above equation, we write values of A,B as:
A=49(loga),B=13(logb)A=\dfrac{-4}{9}\left( \log a \right),\,B=\dfrac{1}{3}\left( logb \right)
By the equation before dividing into terms we can say:
Required expression= A+B
By substituting the values of A, B we get:
log(a4b39)=49loga+13logb\log \left( \sqrt[9]{{{a}^{-4}}{{b}^{3}}} \right)=\dfrac{-4}{9}\log a+\dfrac{1}{3}\log b
By taking least common multiple we can term the equation:
log(a4b39)=49loga+13logb\log \left( \sqrt[9]{{{a}^{-4}}{{b}^{3}}} \right)=\dfrac{-4}{9}\log a+\dfrac{1}{3}\log b.

Note: We proceed with using the identity of logarithm to simplify the expression. Be careful while bringing exponent out from the logarithm. If you don’t care of minus signs in the exponent you might lead to the wrong answer.