Question

The value of ${\log _3}4.{\log _4}5.{\log _5}6.{\log _6}7.{\log _7}8.{\log _8}9$ is
A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

Hint : We are asked to find the value of the given expression. We can observe that there are logarithms in the question. For simplification use the change of base formula of logarithm and apply this formula to each of the terms of the expression separately and then multiply the term to get the answer

** Complete step-by-step answer** :
Given, the expression ${\log _3}4.{\log _4}5.{\log _5}6.{\log _6}7.{\log _7}8.{\log _8}9$
We will the change of base formula of logarithm which is,
${\log _b}\left( a \right) = \dfrac{{{{\log }_N}\left( a \right)}}{{{{\log }_N}\left( b \right)}}$
where $N$ is the base and we can take any value for $N$ .
Let $P = {\log _3}4.{\log _4}5.{\log _5}6.{\log _6}7.{\log _7}8.{\log _8}9$ (i)
Now, let us take each term separately
$A = {\log _3}4$ (ii)
$B = {\log _4}5$ (iii)
$C = {\log _5}6$ (iv)
$D = {\log _6}7$ (v)
$E = {\log _7}8$ (vi)
$F = {\log _8}9$ (vii)
Therefore, now $P$ can be written using equations (ii) to (vii) as,
$P = A.B.C.D.E.F$ (viii)
Now, we apply change of base formula for each term taking the base as $10$
For $A = {\log _3}4$ , we get
$A = {\log _3}4 = \dfrac{{{{\log }_{10}}4}}{{{{\log }_{10}}3}}$
For $B = {\log _4}5$ , we get
$B = {\log _4}5 = \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}4}}$
For $C = {\log _5}6$ , we get
$C = {\log _5}6 = \dfrac{{{{\log }_{10}}6}}{{{{\log }_{10}}5}}$
For $D = {\log _6}7$ , we get
$D = {\log _6}7 = \dfrac{{{{\log }_{10}}7}}{{{{\log }_{10}}6}}$
For $E = {\log _7}8$ , we get
$E = {\log _7}8 = \dfrac{{{{\log }_{10}}8}}{{{{\log }_{10}}7}}$
For $F = {\log _8}9$ , we get
$F = {\log _8}9 = \dfrac{{{{\log }_{10}}9}}{{{{\log }_{10}}8}}$
Now, putting the values of $A,\,B,\,C,\,D,\,E$ and $F$ in equation (viii), we get
$P = \dfrac{{{{\log }_{10}}4}}{{{{\log }_{10}}3}}.\dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}4}}.\dfrac{{{{\log }_{10}}6}}{{{{\log }_{10}}5}}.\dfrac{{{{\log }_{10}}7}}{{{{\log }_{10}}6}}.\dfrac{{{{\log }_{10}}8}}{{{{\log }_{10}}7}}.\dfrac{{{{\log }_{10}}9}}{{{{\log }_{10}}8}}$
$\Rightarrow P = \dfrac{1}{{{{\log }_{10}}3}}.\dfrac{{{{\log }_{10}}9}}{1}$
$\Rightarrow P = \dfrac{{{{\log }_{10}}9}}{{{{\log }_{10}}3}}$
$\Rightarrow P = \dfrac{{{{\log }_{10}}{3^2}}}{{{{\log }_{10}}3}}$ (ix)
We have a formula of logarithm as,
$\log \left( {{a^b}} \right) = b\log a$
Using this formula for ${\log _{10}}{3^2}$ we have,
${\log _{10}}{3^2} = 2{\log _{10}}3$
Putting this value in equation (ix) we get,
$P = \dfrac{{2{{\log }_{10}}3}}{{{{\log }_{10}}3}}$
$\Rightarrow P = 2$
Therefore, the value of the given expression is $2$ .
So, the correct answer is “2.

Note : Remember the basic rules or formula used in logarithm as this will help you to simplify a problem and get an answer easily. We can also solve problems involving logarithms using a calculator or log table but if we know the basic formulas then we can easily solve the problem without using a calculator or log table.