Question

Find the value of $ \dfrac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2} $ .

Answer 1

Hint: Here, we will use some of the properties of logarithm having base 10 i.e. given as $ \log {{m}^{n}}=n\log m $ , $ \log mn=\log m+\log n $ , $ \log \dfrac{m}{n}=\log m-\log n $ , $ {{\log }_{m}}m=1 $ . Then we will simplify the terms in the given property forms like $ \sqrt{27}={{\left( 27 \right)}^{\dfrac{1}{2}}} $ and so on. Thus, on putting values and solving properly we will get our answer.

Complete step-by-step answer:
We have some properties to solve the terms of logarithm to base 10 as given below.
(1) $ \log {{m}^{n}}=n\log m $
(2) $ \log mn=\log m+\log n $
(3) $ \log \dfrac{m}{n}=\log m-\log n $
(4) $ {{\log }_{m}}m=1 $
Here, we will first convert the terms into the above form. So, we can write it as $ \sqrt{27}={{\left( 27 \right)}^{\dfrac{1}{2}}} $ , $ \sqrt{1000}={{\left( 1000 \right)}^{\dfrac{1}{2}}} $ , $ 8={{2}^{3}} $ , $ 1.2=\dfrac{12}{10} $
Now, on putting this in original equation, we get as
 $ \dfrac{\log {{\left( 27 \right)}^{\dfrac{1}{2}}}+\log {{2}^{3}}-\log {{\left( 1000 \right)}^{\dfrac{1}{2}}}}{\log \dfrac{12}{10}} $
So, here we will be using above properties and on solving, we get as
 $ \log {{\left( 27 \right)}^{\dfrac{1}{2}}}=\dfrac{1}{2}\log 27 $ , $ \log {{2}^{3}}=3\log 2 $ , $ \log {{\left( 1000 \right)}^{\dfrac{1}{2}}}=\dfrac{1}{2}\log 1000 $ , $ \log \dfrac{12}{10}=\log 12-\log 10 $ .
On substituting these values, we get as
 $ \dfrac{\dfrac{1}{2}\log 27+3\log 2-\dfrac{1}{2}\log 1000}{\log 12-\log 10} $
Now, further simplifying the terms we can write $ \log 27=\log {{\left( 3 \right)}^{3}} $ , $ \log 1000=\log {{\left( 10 \right)}^{3}} $ , $ \log 12=\log \left( 4\times 3 \right)=\log {{\left( 2 \right)}^{2}}+\log 3 $ , $ {{\log }_{10}}10=1 $
Thus, on putting values we get as
 $ \dfrac{\dfrac{1}{2}\log {{\left( 3 \right)}^{3}}+3\log 2-\dfrac{1}{2}\log {{\left( 10 \right)}^{3}}}{\log {{\left( 2 \right)}^{2}}+\log 3-1} $
On using property (1), we get as
 $ \dfrac{\dfrac{3}{2}\log \left( 3 \right)+3\log 2-\dfrac{3}{2}\log \left( 10 \right)}{2\log \left( 2 \right)+\log 3-1} $
We should know the basic values like $ {{\log }_{10}}3=0.477 $ , $ {{\log }_{10}}2=0.301 $ . So, we get equation as
 $ \dfrac{\dfrac{3}{2}\times 0.477+3\times 0.301-\dfrac{3}{2}\times 1}{2\times 0.301+0.477-1} $
On further solving, we get as
 $ \dfrac{0.7155+0.903-1.5}{0.602+0.477-1} $
 $ \dfrac{0.1185}{0.079} $
Thus, on solving we get answer as
 $ \dfrac{0.1185}{0.079}=1.5 $
Thus, the value of $ \dfrac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2} $ is 1.5 or $ \dfrac{3}{2} $ .

Note: Do not take here base as e i.e. natural log or any other base. If not given any specific base in question, then we will assume log base 10 only. Then only all the properties will obey its properties otherwise the answer will be totally changed and will be wrong. So, do not make this mistake.