Question

we have an expression as ${\log _{12}}27 = a$. So, find the value of ${\log _6}16$ in ‘$a$’ form?

Answer 1

We will use the logarithmic change of base formula for ${\log _{12}}27 = a$ and ${\log _6}16$ . Using change of base formula for ${\log _{12}}27 = a$ , we will find the log of a number with changed base in terms of ‘$a$’ and substitute in ${\log _6}16$ to get the final answer.

Complete step-by-step solution:
Given that: ${\log _{12}}27 = a$
The change of base formula states that,
${\log _q}p = \dfrac{{{{\log }_r}p}}{{{{\log }_r}q}}$
By using this change of base formula, we will modify the equation ${\log _{12}}27 = a$ to base $10$ .
${\log _{12}}27 = a$
${\log _{12}}27 = \dfrac{{{{\log }_{10}}27}}{{{{\log }_{10}}12}} = a$
We can write $27$ as $3 \times 3 \times 3$ and $12$ as $2 \times 2 \times 3$ ,
$\therefore {\log _{12}}27 = \dfrac{{{{\log }_{10}}3 \times 3 \times 3}}{{{{\log }_{10}}2 \times 2 \times 3}} = a$
Substituting $3 \times 3 \times 3 = {3^3}$ and $2 \times 2 \times 3 = {2^2} \times 3$in the above equation,
$\therefore {\log _{12}}27 = \dfrac{{{{\log }_{10}}{3^3}}}{{{{\log }_{10}}({2^2} \times 3)}} = a$
By using the basic logarithmic formula which states that,
${\log _n}(a \times b) = {\log _n}a + {\log _n}b$
We can write ${\log _{12}}27 = \dfrac{{{{\log }_{10}}{3^3}}}{{{{\log }_{10}}({2^2} \times 3)}} = a$ as,
${\log _{12}}27 = \dfrac{{{{\log }_{10}}{3^3}}}{{{{\log }_{10}}{2^2} + {{\log }_{10}}3}} = a$
By using the basic logarithmic formula which states that,
${\log _b}{a^n} = n{\log _b}a$
We write ${\log _{12}}27 = \dfrac{{{{\log }_{10}}{3^3}}}{{{{\log }_{10}}{2^2} + {{\log }_{10}}3}} = a$ as,
$\Rightarrow {\log _{12}}27 = \dfrac{{3{{\log }_{10}}3}}{{2{{\log }_{10}}2 + {{\log }_{10}}3}} = a$
$\Rightarrow \dfrac{{3{{\log }_{10}}3}}{{2{{\log }_{10}}2 + {{\log }_{10}}3}} = a$
Rearranging the terms in the above equation as,
$\dfrac{{2{{\log }_{10}}2 + {{\log }_{10}}3}}{{3{{\log }_{10}}3}} = \dfrac{1}{a}$
Splitting the addition term in the above equation,
$\dfrac{{2{{\log }_{10}}2}}{{3{{\log }_{10}}3}} + \dfrac{{{{\log }_{10}}3}}{{3{{\log }_{10}}3}} = \dfrac{1}{a}$
Solving this equation, we get,
$\dfrac{{2{{\log }_{10}}2}}{{3{{\log }_{10}}3}} + \dfrac{1}{3} = \dfrac{1}{a}$
Taking the $\dfrac{1}{3}$ term to the right-hand side,
$\dfrac{{2{{\log }_{10}}2}}{{3{{\log }_{10}}3}} = \dfrac{1}{a} - \dfrac{1}{3}$
We can write this equation as ,
$\dfrac{2}{3} \times \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}} = \dfrac{{3 - a}}{{a \times 3}}$
Taking the $\dfrac{2}{3}$ term to the right-hand side,
$\dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}} = \dfrac{3}{2} \times \dfrac{{3 - a}}{{a \times 3}}$
$\Rightarrow \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}} = \dfrac{{3(3 - a)}}{{2 \times a \times 3}}$
$\Rightarrow \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}} = \dfrac{{(3 - a)}}{{2a}}$
$\Rightarrow \dfrac{{2a}}{{(3 - a)}}{\log _{10}}2 = {\log _{10}}3$ ….. (1)
Now consider,
${\log _6}16$
Applying the same change of base formula,
${\log _q}p = \dfrac{{{{\log }_r}p}}{{{{\log }_r}q}}$
${\log _6}16 = \dfrac{{{{\log }_{10}}16}}{{{{\log }_{10}}6}}$
We can write $16 = 2 \times 2 \times 2 \times 2$ and $6 = 2 \times 3$ ,
${\log _6}16 = \dfrac{{{{\log }_{10}}2 \times 2 \times 2 \times 2}}{{{{\log }_{10}}2 \times 3}}$
Substituting $2 \times 2 \times 2 \times 2 = {2^4}$,
${\log _6}16 = \dfrac{{{{\log }_{10}}{2^4}}}{{{{\log }_{10}}2 \times 3}}$
By using the basic logarithmic formula which states that,
${\log _n}(a \times b) = {\log _n}a + {\log _n}b$
We can write ${\log _6}16 = \dfrac{{{{\log }_{10}}{2^4}}}{{{{\log }_{10}}2 \times 3}}$ as,
${\log _6}16 = \dfrac{{{{\log }_{10}}{2^4}}}{{{{\log }_{10}}2 + {{\log }_{10}}3}}$
By using the basic logarithmic formula which states that,
${\log _b}{a^n} = n{\log _b}a$
${\log _6}16 = \dfrac{{4{{\log }_{10}}2}}{{{{\log }_{10}}2 + {{\log }_{10}}3}}$
Substituting the value of ${\log _{10}}3$ from equation (1),
${\log _6}16 = \dfrac{{4{{\log }_{10}}2}}{{{{\log }_{10}}2 + \dfrac{{2a}}{{(3 - a)}}{{\log }_{10}}2}}$
Taking ${\log _{10}}2$ common from the denominator,
${\log _6}16 = \dfrac{{4{{\log }_{10}}2}}{{{{\log }_{10}}2\left( {1 + \dfrac{{2a}}{{(3 - a)}}} \right)}}$
Cancelling out ${\log _{10}}2$ from numerator and denominator we get,
${\log _6}16 = \dfrac{4}{{\left( {1 + \dfrac{{2a}}{{(3 - a)}}} \right)}}$
Simplifying this equation further,
${\log _6}16 = \dfrac{4}{{\left( {\dfrac{{3 - a + 2a}}{{(3 - a)}}} \right)}}$
$\Rightarrow {\log _6}16 = \dfrac{{4(3 - a)}}{{3 - a + 2a}}$
$\Rightarrow {\log _6}16 = \dfrac{{4(3 - a)}}{{(3 + a)}}$
The value of ${\log _6}16$ in ‘$a$’ form is ${\log _6}16 = \dfrac{{4(3 - a)}}{{(3 + a)}}$ .

Note: The change of base method allows rewriting the logarithm in phrases of any other base log. change of base formulation is used within the evaluation of log and has every other base than $10$.