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Question

Question: The value of \(\log 5+\log 8-2\log 2\) is equal to: (a). 1 (b). 8 (c). 4 (d). 2...

The value of log5+log82log2\log 5+\log 8-2\log 2 is equal to:
(a). 1
(b). 8
(c). 4
(d). 2

Explanation

Solution

Hint: The easiest way to approach this problem and simplify the given question is by using the logarithmic property logM+logN=log(MN)\log M+\log N=\log \left( MN \right) and MlogN=logNMM\log N=\log {{N}^{M}} . Further, since the base of logarithm is not given, it is assumed to be 10.

Complete step-by-step answer:

There are four basic rules of logarithms as given below:-
logb(mn)=logbm+logbn{{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n
logb(mn)=logbmlogbn{{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n
logb(mn)=nlogbm{{\log }_{b}}\left( {{m}^{n}} \right)=n{{\log }_{b}}m
logbm=logamlogablo{{g}_{b}}m=\dfrac{{{\log }_{a}}m}{{{\log }_{a}}b}

Let us now consider the given question,

We know that, logM+logN=log(MN)\log M+\log N=\log \left( MN \right) and MlogN=logNMM\log N=\log {{N}^{M}} .

Therefore, in case of this problem, we have,
log5 + log8 -2log2

Let us now use the property that logb(mn)=logbm+logbn{{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n to simplify log5+log8\log 5+\log 8 and use logb(mn)=nlogbm{{\log }_{b}}\left( {{m}^{n}} \right)=n{{\log }_{b}}m to simplify 2log22\log 2

=log5×8log22=\log 5\times 8-\log {{2}^{2}}
=log40log4=\log 40-\log 4
Here we are using the property that logb(mn)=logbmlogbn{{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n to simplify log40log4\log 40-\log 4
Thus, we have,
=log404=\log \dfrac{40}{4}
=log10=\log 10
We know that log 10 is actually log1010{{\log }_{10}}10 .
Let us equate log1010=n{{\log }_{10}}10=n
10n=10\Rightarrow {{10}^{n}}=10
n=1\Rightarrow n=1

Hence, log10=1\log 10=1

Therefore, the final answer is option A.

Note: When we approach this question a lot of us make the mistake of thinking that we need to use the logarithmic tables to solve the question but, on the contrary, using logarithmic properties is much more efficient as it is useful for solving complex logarithmic problems as well.