Question

How do you find the exact value of \[{{\log }_{4}}2+{{\log }_{4}}32\]?

Answer 1

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 \[{{\log }_{4}}2+{{\log }_{4}}32\].
Here, we have base 4 rather than 10. We know that, whenever we have 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 2 to the base 4 as the common logarithm of 2 divided by the common logarithm of 4. That is, we get
\[\begin{align}
  & \Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=\dfrac{{{\log }_{10}}2}{{{\log }_{10}}4}+\dfrac{5{{\log }_{10}}2}{{{\log }_{10}}4} \\
 & \Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=3 \\
 & \Rightarrow {{\log }_{10}}4={{\log }_{10}}{{2}^{2}}=2{{\log }_{10}}2 \\
 & x={{10}^{y}}x\underline{<0} \\
\end{align}\]
\[\Rightarrow {{\log }_{4}}2=\dfrac{{{\log }_{10}}2}{{{\log }_{10}}4}\] --------(1)
we can express the common logarithm of 32 to the base 4 as the common logarithm of 32 divided by the common logarithm of 4. That is, we get
\[\Rightarrow {{\log }_{4}}32=\dfrac{{{\log }_{10}}32}{{{\log }_{10}}4}\] ---------(2)
We know that 32 can be written as the fifth power of 2. That is, \[32=2\times 2\times 2\times 2\times 2={{2}^{5}}\].
Therefore, we can rewrite equation 2 as
\[\Rightarrow {{\log }_{4}}32=\dfrac{{{\log }_{10}}32}{{{\log }_{10}}4}=\dfrac{{{\log }_{10}}{{2}^{5}}}{{{\log }_{10}}4}\] ----------(3)
Using the property \[{{\log }_{b}}{{p}^{r}}=r{{\log }_{b}}p\], we can write the logarithm of \[{{2}^{5}}\]as 5 times logarithm of 2. That is, we can rewrite the equation 3 as
\[\Rightarrow {{\log }_{4}}32=\dfrac{5{{\log }_{10}}2}{{{\log }_{10}}4}\] ---------(4)
Now we add equations 1 and 4 then we get
\[\begin{align}
  & \Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=\dfrac{{{\log }_{10}}2}{{{\log }_{10}}4}+\dfrac{5{{\log }_{10}}2}{{{\log }_{10}}4} \\
 & \Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=\dfrac{6{{\log }_{10}}2}{{{\log }_{10}}4} \\
\end{align}\] --------(5)
We know that 4 can be written as the square of 2. That is, \[4={{2}^{2}}\].
Using the property \[{{\log }_{b}}{{p}^{r}}=r{{\log }_{b}}p\], we can write the logarithm of a square of 2 as 2 times logarithm of 2.
\[\Rightarrow {{\log }_{10}}4={{\log }_{10}}{{2}^{2}}=2{{\log }_{10}}2\] --------(6)
On substituting equation 6 in 5, we get
\[\Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=\dfrac{6{{\log }_{10}}2}{2{{\log }_{10}}2}\] ---------(7)
Since \[{{\log }_{10}}2\] is common in numerator and denominator. So, the equation will be
\[\Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=\dfrac{6}{2}\]
\[\Rightarrow {{\log }_{4}}2+{{\log }_{4}}32=3\]
Therefore, the exact value of \[{{\log }_{4}}2+{{\log }_{4}}32\] is 3.

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 \[x\underline{<0}\]. We must avoid calculation mistakes to get the correct result.