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Question

Question: Show that \[\dfrac{1}{2}\log 9 + 2\log 6 + \dfrac{1}{4}\log 81 - \log 12 = 3\log 3\]...

Show that 12log9+2log6+14log81log12=3log3\dfrac{1}{2}\log 9 + 2\log 6 + \dfrac{1}{4}\log 81 - \log 12 = 3\log 3

Explanation

Solution

Hint : Here the given function is a logarithm function it can be defined as logarithmic functions are the inverses of exponential functions. For the given log function we can assume base value 10, by using some of the Basic Properties of logarithmic functions. And by further simplification we get a required solution.

Complete step-by-step answer :
The function from positive real numbers to real numbers to real numbers is defined as logb:R+Rlogb(x)=y{\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y , if by=x{b^y} = x , is called logarithmic function or the logarithm function is the inverse form of exponential function.
There are some basic logarithms properties
1. product rule :- log(mn)=logm+logn\log \left( {mn} \right) = \log m + \log n
2. Quotient rule :- log(mn)=logmlogn\log \left( {\dfrac{m}{n}} \right) = \log m - \log n
3. Power rule :- log(mn)=n.logm\log \left( {{m^n}} \right) = n.\log m
Now, Consider the given logarithm function
12log9+2log6+14log81log12=3log3\dfrac{1}{2}\log 9 + 2\log 6 + \dfrac{1}{4}\log 81 - \log 12 = 3\log 3
Now we will consider the LHS and we have to show that the LHS is equal to the RHS.
So now we will consider the LHS
12log9+2log6+14log81log12\Rightarrow \dfrac{1}{2}\log 9 + 2\log 6 + \dfrac{1}{4}\log 81 - \log 12
Here in the first term we write 9 as 32{3^2} , for the second term we apply the product rule, in the third term we write 81 as 34{3^4} , so the above inequality is written as
12log32+log62+14log34log12\Rightarrow \dfrac{1}{2}\log {3^2} + \log {6^2} + \dfrac{1}{4}\log {3^4} - \log 12
For the first term and the third term we apply the product rule and it is rewritten as
122log3+log36+144log3log12\Rightarrow \dfrac{1}{2}2\log 3 + \log 36 + \dfrac{1}{4}4\log 3 - \log 12
on simplifying we have
log3+log3+log36log12\Rightarrow \log 3 + \log 3 + \log 36 - \log 12
For the third and the last term we will use the quotient rule and it is written as
log3+log3+log3612\Rightarrow \log 3 + \log 3 + \log \dfrac{{36}}{{12}}
On simplifying we have
log3+log3+log3\Rightarrow \log 3 + \log 3 + \log 3
Therefore we have
3log3\Rightarrow 3\log 3
RHS\Rightarrow RHS
Hence showed

Note : The question contains the log terms we must know the logarithmic properties which are the standard properties. By applying properties we can solve the question in an easy manner. We apply the formula or properties of the logarithmic functions. where it is necessary. Hence, we obtain the desired result.