
How do you evaluate \[{{\log }_{27}}3\]?
Answer
546k+ views
Hint: Convert the base of the given logarithmic function, i.e., 27 into the exponential form having base 3. Now, apply the logarithmic formula given as: - \[{{\log }_{\left( {{a}^{b}} \right)}}m=\left( \dfrac{1}{b} \right){{\log }_{a}}m\] to simplify the expression. Now, use the identity: - \[{{\log }_{n}}n=1\] to get the answer. Here, ‘n’ denotes the same argument and base of the logarithmic function and n > 0; n \[\ne \] 1.
Complete step by step answer:
Here, we have been provided with the logarithmic expression: - \[{{\log }_{27}}3\] and we are asked to evaluate it. That means we need to find its value.
Now, the given logarithmic function is a common log expression. That means the base value of the log 10. Let us assume the value of this expression as ‘E’. So, we have,
\[\Rightarrow E={{\log }_{27}}3\]
Converting the base, i.e., 27 into exponential form, we get,
\[\Rightarrow E={{\log }_{{{3}^{3}}}}3\]
Using the logarithmic formula: - \[{{\log }_{\left( {{a}^{b}} \right)}}m=\left( \dfrac{1}{b} \right){{\log }_{a}}m\], we get,
\[\Rightarrow E=\left( \dfrac{1}{3} \right){{\log }_{3}}3\]
Now, since the base and argument of the logarithmic expression is the same and we know that \[{{\log }_{n}}n=1\], where n > 0, n \[\ne \] 1. So, we have,
\[\begin{align}
& \Rightarrow E=\dfrac{1}{3}\times 1 \\
& \Rightarrow E=\dfrac{1}{3} \\
\end{align}\]
Hence, \[\dfrac{1}{3}\] is the simplified form of \[{{\log }_{27}}3\].
Note:
One must know the difference between common log and natural log to solve the above question. Common log has base 10 and natural log has base e. Natural log is denoted by ln. You must remember the basic formulas of logarithm like: - \[\log m+\log n=\log \left( mn \right)\], \[\log m-\log n=\log \dfrac{m}{n}\], \[\log {{m}^{n}}=n\log m\], \[{{\log }_{\left( {{a}^{b}} \right)}}m=\left( \dfrac{1}{b} \right){{\log }_{a}}m\]. Note that we can also solve the question by converting the logarithmic form of the equation into exponential form by using the formula: - \[x={{\log }_{a}}m\] then \[m={{a}^{x}}\]. In the next step we will write: - \[{{3}^{1}}={{27}^{x}}\Rightarrow {{3}^{1}}={{3}^{3x}}\] and then compare the exponents (1) and (3x) to solve for the value of x.
Complete step by step answer:
Here, we have been provided with the logarithmic expression: - \[{{\log }_{27}}3\] and we are asked to evaluate it. That means we need to find its value.
Now, the given logarithmic function is a common log expression. That means the base value of the log 10. Let us assume the value of this expression as ‘E’. So, we have,
\[\Rightarrow E={{\log }_{27}}3\]
Converting the base, i.e., 27 into exponential form, we get,
\[\Rightarrow E={{\log }_{{{3}^{3}}}}3\]
Using the logarithmic formula: - \[{{\log }_{\left( {{a}^{b}} \right)}}m=\left( \dfrac{1}{b} \right){{\log }_{a}}m\], we get,
\[\Rightarrow E=\left( \dfrac{1}{3} \right){{\log }_{3}}3\]
Now, since the base and argument of the logarithmic expression is the same and we know that \[{{\log }_{n}}n=1\], where n > 0, n \[\ne \] 1. So, we have,
\[\begin{align}
& \Rightarrow E=\dfrac{1}{3}\times 1 \\
& \Rightarrow E=\dfrac{1}{3} \\
\end{align}\]
Hence, \[\dfrac{1}{3}\] is the simplified form of \[{{\log }_{27}}3\].
Note:
One must know the difference between common log and natural log to solve the above question. Common log has base 10 and natural log has base e. Natural log is denoted by ln. You must remember the basic formulas of logarithm like: - \[\log m+\log n=\log \left( mn \right)\], \[\log m-\log n=\log \dfrac{m}{n}\], \[\log {{m}^{n}}=n\log m\], \[{{\log }_{\left( {{a}^{b}} \right)}}m=\left( \dfrac{1}{b} \right){{\log }_{a}}m\]. Note that we can also solve the question by converting the logarithmic form of the equation into exponential form by using the formula: - \[x={{\log }_{a}}m\] then \[m={{a}^{x}}\]. In the next step we will write: - \[{{3}^{1}}={{27}^{x}}\Rightarrow {{3}^{1}}={{3}^{3x}}\] and then compare the exponents (1) and (3x) to solve for the value of x.
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