Question

Express each of the following as a single logarithm:
\[\dfrac{1}{2}\log 36 + 2\log 8 - \log 1.5\]

Answer 1

Hint: In order to solve this question first, we assume the variable that is equal to the given expression. Then we have to use the properties of the logarithm. First, eliminate the coefficient of the logarithm by taking all of them to the powers. Then make all the numbers that are in decimal format and in logarithm function to the rational number. Then use the properties of the logarithm to merge two numbers which are in addition after solving this we will get the answer in a single logarithm.

Complete step-by-step answer:
Let \[x = \dfrac{1}{2}\log 36 + 2\log 8 - \log 1.5\]
Using the property of logarithm:
\[a\log b = \log {b^a}\]
\[ = \log {36^{\dfrac{1}{2}}} + \log {8^2} - \log 1.5\]
On solving the power
\[ = \log 6 + \log 64 - \log 1.5\]
On making all to rational number
\[ = \log 6 + \log 64 - \log \dfrac{{15}}{{10}}\]
Using the property of logarithm:
\[\log \dfrac{a}{b} = \log a - \log b\]
\[ = \log 6 + \log 64 - (\log 15 - \log 10)\]
On further solving
\[ = \log 6 + \log 64 - \log 15 + \log 10\]
Now using the property of logarithm
\[\log a + \log b + \log c = \log a \times b \times c\]
\[ = \log \left( {6 \times 64 \times 10} \right) - \log 15\]
On further solving:
\[ = \log \left( {3840} \right) - \log 15\]
Now using the same property again
\[\log \dfrac{a}{b} = \log a - \log b\]
\[ = \log \left( {\dfrac{{3840}}{{15}}} \right)\]
On calculating
\[ \Rightarrow x = \log 256\]
Final answer:
The value of \[x\]is
\[ \Rightarrow x = \log \left( {256} \right)\]

Note: In solving this question you may commit a mistake in applying the properties on the logarithm functions. In order to solve this type of question, you must know all the properties of the logarithm function and try to apply all those whenever required. Properties used in order to solve this question are.
\[a\log b = \log {b^a}\],
\[\log \dfrac{a}{b} = \log a - \log b\],
\[\log a + \log b + \log c = \log a \times b \times c\], and
\[\log \dfrac{a}{b} = \log a - \log b\] there are many properties other than these so remember all to solve difficult questions. Students are confused in applying the properties of logarithm and may commit mistakes in the calculations also so don’t make any mistakes in calculations also.