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

Last updated date: 20th Jun 2024
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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.