Question

Expand log \[\sqrt[{12}]{{{a^3}{b^2}{c^4}}}\]as log a, log b, log c.

Answer 1

Hint: We have given that log \[\sqrt[{12}]{{{a^3}{b^2}{c^4}}}\]we expand it is the form of log a, log b, log c. So firstly have to write the 12th root of \[{a^3}{b^4}{c^2}\]in\[{\left( {{a^3}{b^2}{c^4}} \right)^{1{/_{12}}}}\]. Then we apply the property of logarithm. This properly helps to expand the log again we apply the property of logarithm on the product of function. This will separate log\[{a^3}\], log\[{b^2}\]and log\[{c^4}\]. Again applying the proper log \[{m^n}\]we will get the required result.

Complete step-by-step solution:
Let consider the given log function
\[ \Rightarrow \,\,\,I = \log \sqrt[{12}]{{{a^3}{b^2}{c^4}}}\]------------(1)
Here we have to expand it in log a, log b, log c.
Rewrite the equation (1) in exponent form, then
\[ \Rightarrow \,\,\,I = \log {\left( {{a^3}{b^2}{c^4}} \right)^{\dfrac{1}{{12}}}}\]----------(2)
Now we apply the properties of logarithm function. We know that \[\log {m^n} = {\text{ }}n\log m\].
Then equation (2) becomes
\[ \Rightarrow \,\,\,I = \dfrac{1}{{12}}\log \left( {{a^3}{b^2}{c^4}} \right)\]-------(3)
Also we know that product property of logarithm function i.e., \[\log \left( {mn} \right) = \log m + \log n\]
Then equation (3) becomes
\[ \Rightarrow \,\,\,\,\,I = \dfrac{1}{{12}}\left[ {\log {a^3} + {\text{ log}}{b^2} + {\text{ log}}{c^4}} \right]\]
Multiply \[\dfrac{1}{{12}}\] to each term inside the parenthesis, then
\[ \Rightarrow \,\,\,\,\,I = \dfrac{1}{{12}}\log {a^3} + \dfrac{1}{{12}}{\text{log}}{b^2} + \dfrac{1}{{12}}{\text{log}}{c^4}\]----------(4)
 Now again by the properties of logarithm \[\log {m^n} = {\text{ }}n\log m\]
Then equation (4) becomes
\[ \Rightarrow \,\,\,\,I = \dfrac{1}{{12}} \times 3\log a{\text{ + }}\dfrac{1}{{12}} \times 2\log b + \dfrac{1}{{12}} \times 4\log c\]
On simplification, we get
\[ \Rightarrow \,\,\,\,I = \dfrac{1}{4}\log a{\text{ + }}\dfrac{1}{6}\log b + \dfrac{1}{3}\log c\]
Hence, the expand form of the function \[\log \sqrt[{12}]{{{a^3}{b^2}{c^4}}}\] is \[\,\,I = \dfrac{1}{4}\log a{\text{ + }}\dfrac{1}{6}\log b + \dfrac{1}{3}\log c\].

Note: In mathematics logarithm is the inverse function to exponentiation that means the logarithm of a given number x is the exponent to which another fixed number. The base b must raise, to produce that number x. In the simplest case the logarithm counts the number assurances of the same factor in repeated multiplication. EX: \[10 \times 10 \times 10 = {10^3}\]the logarithm base \[10\] of \[1000\] is 3 or \[{\log _{10}}\] \[\left( {1000} \right)\]= 3. The logarithm of x base b is as \[{\log _b}\] (x) or without parentheses \[{\log _b}\]x…or without explicit base log x. Most generally, exponentiation allows any positive real number b and x where b is not equal to 1 is always unique real number y.