Question

log12 to base 7=a, log 24 to base12=b . Find log 168 to base 54​

Answer 1

To find: $\log_{54}168$

Here, $\log_{7}12=a$ and $\log_{12}24=b$

Using the property $\log_{a} b=\frac{1}{\log_{b}a}$

$\log_{12}7=\frac{1}{a}$

Thus,

$\log_{12}7+\log_{12}24=\frac{1}{a}+b$

Using the property $\log (mn)=\log m+\log n$

$\log_{12}168=\frac{1}{a}+b$     ...... (1)

Now, using the property $\log_{a}b=\frac{\log_{x}b}{\log_{x}a}$,

$\log_{54}168=\frac{\log_{12}168}{\log_{12}54}$

From equation (1),

$\log_{54}168=\frac{b+\frac{1}{a}}{\log_{12}54}$   ...... (2)

Again, $\log_{12}54=\log_{12}(3^3\times 2)$

                       $=\log_{12}3^3+\log_{12}2$

                       $=3\log_{12}3+\log_{12}2$                (∵ $\log a^b=b\log a$  )

Solve $\log_{12}24=b$ as follows:

        $\log_{12}(3\times 2^3)=b$

$\log_{12}(3)+3\log_{12} 2=b$

                  $\log_{12} 3=b-3\log_{12}(2)$

Thus,

$\log_{12}54=3b-9\log_{12}(2)+\log_{12}2$

            $=3b-8\log_{12}(2)$

Also, $\log_{12}24=b$

     $\log_{12}(12\times 2)=b$

$\log_{12}12+\log_{12}2=b$          

         $1+\log_{12}2=b$                   (∵ $\log_{a} a=1$)

        $\implies \log_{12}2=b-1$

So,

$\log_{12}54=3b-8(b-1)$

            $=3b-8b+8$

            $=8-5b$

That is, $\log_{12}54=8-5b$

Substitute this value in equation (2),

$\log_{54}168=\frac{b+\frac{1}{a}}{8-5b}$

             $=\frac{ab+1}{8a-5ab}$

Therefore, $\log_{54}168=\frac{ab+1}{8a-5ab}$