How do you evaluate \[{{\log }_{3}}4\]?
Answer
630.9k+ views
Hint: A common logarithm is the logarithm of base 10. To get the logarithm of a number \[n\], find the number \[x\] that when the base is raised to that power, the resulting value is \[n\]. We can find common logarithm of a number with a base other than 10 using the property \[{{\log }_{b}}a=\dfrac{\log a}{\log b}\]. While finding \[{{\log }_{b}}a\], if \[a={{p}^{r}}\], then we can write the power \[r\] in front of \[\log a\]. That is, \[{{\log }_{b}}{{p}^{r}}=r{{\log }_{b}}p\].
Complete step by step answer:
As per the given question, we have to find the common logarithm of 4 to the base 3. That is, we need to evaluate \[{{\log }_{3}}4\].
Here, we have base 3 rather than 10. We know that, whenever have a different base other than 10 for common logarithm, we can use the property \[{{\log }_{b}}a=\dfrac{\log a}{\log b}\].
Therefore, we can express the common logarithm of 4 to the base 3 as the common logarithm of 4 divided by the common logarithm of 3. That is, we get
\[\Rightarrow {{\log }_{3}}4=\dfrac{{{\log }_{10}}4}{{{\log }_{10}}3}\] ---------(1)
We know that 4 can be written as the square of 2. That is, \[4={{2}^{2}}\].
Therefore, we can rewrite equation 1 as
\[\Rightarrow {{\log }_{3}}4=\dfrac{{{\log }_{10}}4}{{{\log }_{10}}3}=\dfrac{{{\log }_{10}}{{2}^{2}}}{{{\log }_{10}}3}\] ----------(2)
Using the property \[{{\log }_{b}}{{p}^{r}}=r{{\log }_{b}}p\], we can write logarithm of square of 2 as 2 times logarithm of 2. That is, we can rewrite the equation 2 as
\[\Rightarrow {{\log }_{3}}4=\dfrac{2{{\log }_{10}}2}{{{\log }_{10}}3}\] --------(3)
We know that logarithm of 2 to the base 10 is equal to 0.301 and logarithm of 3 to the base 10 is equal to 0.4771. So, by substituting \[{{\log }_{10}}2=0.301\] and \[{{\log }_{10}}3=0.4771\] in the equation 3, we get
\[\Rightarrow {{\log }_{3}}4=\dfrac{2\times 0.301}{0.4771}\]
On multiplying 0.301 with 2 and then dividing the value by 0.4771, we get
\[\Rightarrow {{\log }_{3}}4=\dfrac{0.602}{0.4771}=1.2618\]
\[\therefore 1.2618\] is the required value of \[{{\log }_{3}}4\].
Note:
The most common mistake made with the common logarithm is simply forgetting that we are dealing with a logarithmic function. And this function does not exist for the values of x equal to or less than 0. That is, there is no value of y in the equation \[x={{10}^{y}}\] for which \[x\le 0\]. We must avoid calculation mistakes to get the correct result.
Complete step by step answer:
As per the given question, we have to find the common logarithm of 4 to the base 3. That is, we need to evaluate \[{{\log }_{3}}4\].
Here, we have base 3 rather than 10. We know that, whenever have a different base other than 10 for common logarithm, we can use the property \[{{\log }_{b}}a=\dfrac{\log a}{\log b}\].
Therefore, we can express the common logarithm of 4 to the base 3 as the common logarithm of 4 divided by the common logarithm of 3. That is, we get
\[\Rightarrow {{\log }_{3}}4=\dfrac{{{\log }_{10}}4}{{{\log }_{10}}3}\] ---------(1)
We know that 4 can be written as the square of 2. That is, \[4={{2}^{2}}\].
Therefore, we can rewrite equation 1 as
\[\Rightarrow {{\log }_{3}}4=\dfrac{{{\log }_{10}}4}{{{\log }_{10}}3}=\dfrac{{{\log }_{10}}{{2}^{2}}}{{{\log }_{10}}3}\] ----------(2)
Using the property \[{{\log }_{b}}{{p}^{r}}=r{{\log }_{b}}p\], we can write logarithm of square of 2 as 2 times logarithm of 2. That is, we can rewrite the equation 2 as
\[\Rightarrow {{\log }_{3}}4=\dfrac{2{{\log }_{10}}2}{{{\log }_{10}}3}\] --------(3)
We know that logarithm of 2 to the base 10 is equal to 0.301 and logarithm of 3 to the base 10 is equal to 0.4771. So, by substituting \[{{\log }_{10}}2=0.301\] and \[{{\log }_{10}}3=0.4771\] in the equation 3, we get
\[\Rightarrow {{\log }_{3}}4=\dfrac{2\times 0.301}{0.4771}\]
On multiplying 0.301 with 2 and then dividing the value by 0.4771, we get
\[\Rightarrow {{\log }_{3}}4=\dfrac{0.602}{0.4771}=1.2618\]
\[\therefore 1.2618\] is the required value of \[{{\log }_{3}}4\].
Note:
The most common mistake made with the common logarithm is simply forgetting that we are dealing with a logarithmic function. And this function does not exist for the values of x equal to or less than 0. That is, there is no value of y in the equation \[x={{10}^{y}}\] for which \[x\le 0\]. We must avoid calculation mistakes to get the correct result.
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