Question

if log 27 base 12 is equals to a then find log 16 base 6

Answer 1

HEY mate here is your answer.

Hope it helps you.

^_^

Answer 2

Answer:

$\frac{4(3-a)}{3+ a}$ is the required value of $log_{6}16$ .

Step-by-step explanation:

Explanation:

Given , $log_{12}27$ = a

So we need to find the value of $log_{6}16$ .

Step 1:

We have ,$log_{12}27 = a$

⇒$\frac{log27}{log12} = a$                        [ ∴ $log_{a}b = \frac{logb}{loga}$]

⇒$\frac{log3^{3}}{log(2^{2}+ 3) } = a$   ⇒ ${log3^{3}} = a({log(2^{2}.3) })$   [Where log (a.b) = loga + log b]

⇒$3{log3} = a(2log2+ log 3) )$

⇒3log 3 - alog3 = 2alog2

⇒log3 (3 - a ) = 2a log 2   ⇒   $log3 = \frac{2alog2}{3-a}$......(i)

Step 2:

We have , $log_{6}16$  this can be written as  $\frac{log16}{log6}$

⇒$\frac{log2^{4} }{log(2.3)}$ = $\frac{4log2}{log2 + log 3}$ ..........(ii)

Now put the value of log3 from equation(i)  in equation (ii)

⇒$\frac{4log2}{log2 + \frac{2alog2}{3-a}}$ = $\frac{4(3-a)}{3+ a}$.

Final answer :

Hence , $\frac{4(3-a)}{3+ a}$ is the value of $log_{6}16$ .

#SPJ2