Question

If ${\log _{12}}27 = a$, then what is ${\log _6}16$ ?

Answer 1

Hint: We need to find the value of ${\log _6}16$ provided ${\log _{12}}27 = a$
First, we need to analyze the given data so that we can able to solve the problem. We need to solve the given ${\log _{12}}27 = a$ using the power rule, the product rule, and the change of base rule. Using these rules, we will obtain a result. Then we need to start to solve ${\log _6}16$ and we need to apply the obtained result in it.

Formula to be used:
A) The product rule is given below.
\[lo{g_{b\;}}MN = lo{g_b}M + lo{g_{b\;}}N\]
B) The power rule to be applied is as follows.
\[lo{g_{b\;}}{M^p}\; = p{\text{ }}lo{g_b}M\]
C) The change of base rule to be applied is as follows.
\[\;lo{g_{b\;}}\left( x \right) = \dfrac{{log\;\left( x \right)}}{{log\;\left( b \right)}}\]

Complete step-by-step answer:
It is given that ${\log _{12}}27 = a$
We are asked to calculate the value of ${\log _6}16$
Let us consider the given ${\log _{12}}27 = a$
We shall apply the change of base rule \[\;lo{g_{b\;}}\left( x \right) = \dfrac{{log\;\left( x \right)}}{{log\;\left( b \right)}}\]
Thus, we have $\dfrac{{\log 27}}{{\log 12}} = a$
$ \Rightarrow a = \dfrac{{\log 27}}{{\log 12}}$
$ \Rightarrow a = \dfrac{{\log {3^3}}}{{\log 12}}$ (Here we have changed $27$ into ${3^3}$)
$ \Rightarrow a = \dfrac{{3\log 3}}{{\log 12}}$ (Here we have applied \[lo{g_{b\;}}{M^p}\; = p{\text{ }}lo{g_b}M\])
$ \Rightarrow a = \dfrac{{3\log 3}}{{\log \left( {3 \times 4} \right)}}$ (We rewritten $12$ as $3 \times 4$ )
$ \Rightarrow a = \dfrac{{3\log 3}}{{\log 3 + \log 4}}$ (Here we applied \[lo{g_{b\;}}MN = lo{g_b}M + lo{g_{b\;}}N\])
$ \Rightarrow a = \dfrac{{3\log 3}}{{\log 3 + \log {2^2}}}$ (Here we have changed $4$ into ${2^2}$)
$ \Rightarrow a = \dfrac{{3\log 3}}{{\log 3 + 2\log 2}}$ (Here we have applied \[lo{g_{b\;}}{M^p}\; = p{\text{ }}lo{g_b}M\])
$ \Rightarrow a\left( {\log 3 + 2\log 2} \right) = 3\log 3$
$ \Rightarrow a\log 3 + 2a\log 2 = 3\log 3$
$ \Rightarrow 2a\log 2 = 3\log 3 - a\log 3$
$ \Rightarrow 2a\log 2 = \log 3\left( {3 - a} \right)$
$ \Rightarrow \log 2 = \dfrac{{\log 3\left( {3 - a} \right)}}{{2a}}$ ………….$\left( 1 \right)$
Now, we shall consider ${\log _6}16$
We shall apply the change of base rule \[\;lo{g_{b\;}}\left( x \right) = \dfrac{{log\;\left( x \right)}}{{log\;\left( b \right)}}\]
Thus, we have ${\log _6}16 = \dfrac{{\log 16}}{{\log 6}}$
$ \Rightarrow {\log _6}16 = \dfrac{{\log {4^2}}}{{\log 6}}$ (Here we have changed $16$ into ${2^4}$)
$ \Rightarrow {\log _6}16 = \dfrac{{4\log 2}}{{\log 6}}$ (Here we have applied \[lo{g_{b\;}}{M^p}\; = p{\text{ }}lo{g_b}M\])
$ \Rightarrow {\log _6}16 = \dfrac{{4\log 2}}{{\log \left( {2 \times 3} \right)}}$ (We rewritten $6$ as $3 \times 2$ )
$ \Rightarrow {\log _6}16 = \dfrac{{4\log 2}}{{\log 2 + \log 3}}$ (Here we applied \[lo{g_{b\;}}MN = lo{g_b}M + lo{g_{b\;}}N\])
Now, we shall substitute the equation $\left( 1 \right)$ in the above equation.
\[ \Rightarrow {\log _6}16 = \dfrac{{4\dfrac{{\log 3\left( {3 - a} \right)}}{{2a}}}}{{\dfrac{{\log 3\left( {3 - a} \right)}}{{2a}} + \log 3}}\]
\[ \Rightarrow {\log _6}16 = \dfrac{{4\dfrac{{\log 3\left( {3 - a} \right)}}{{2a}}}}{{\dfrac{{\log 3\left( {3 - a} \right) + 2a\log 3}}{{2a}}}}\]
\[ \Rightarrow {\log _6}16 = \dfrac{{4\log 3\left( {3 - a} \right)}}{{\log 3\left( {3 - a} \right) + 2a\log 3}}\]
\[ \Rightarrow {\log _6}16 = \dfrac{{4\log 3\left( {3 - a} \right)}}{{\log 3\left( {\left( {3 - a} \right) + 2a} \right)}}\]
\[ \Rightarrow {\log _6}16 = \dfrac{{4\left( {3 - a} \right)}}{{\left( {\left( {3 - a} \right) + 2a} \right)}}\]
\[ \Rightarrow {\log _6}16 = \dfrac{{12 - 4a}}{{2a - a + 3}}\]
\[ \Rightarrow {\log _6}16 = \dfrac{{12 - 4a}}{{a + 3}}\]
Hence, we found the required answer \[{\log _6}16 = \dfrac{{12 - 4a}}{{a + 3}}\].

Note: While solving the equations containing logarithm, we need to be aware of all the basic rules that can be applied in the equation. The basic rules are the product rule, the quotient rule, the power rule, and the change of base rule.
The quotient rule is \[lo{g_{b\;}}\dfrac{M}{N} = lo{g_b}M - lo{g_{b\;}}N\]