Question

please someone help me with this sum!​

Answer 1

We're given to prove,

$\dfrac{1}{\log_836}+\dfrac{1}{\log_936}+\dfrac{1}{\log_{18}36}=2$

We use,

$1.\ \ \log_ab\ =\ \dfrac{\log b}{\log a}$

$2.\ \ \log a_1+\log a_2+\log a_3+...+\log a_n\ =\ \log(a_1\cdot a_2\cdot a_3\cdot...\cdot a_n)$

or simply,

$\displaystyle 2.\ \ \sum_{k=1}^{n}\log a_k\ =\ \log \left(\prod_{k=1}^{n}a_k\right)$

and,

$3.\ \ \log(a^b)\ =\ b\log a$

Now,

$\begin{aligned}&\textsf{LHS}\\ \\ \Longrightarrow\ \ &\dfrac{1}{\log_836}+\dfrac{1}{\log_936}+\dfrac{1}{\log_{18}36}\\ \\ \Longrightarrow\ \ &\dfrac{1}{\left(\dfrac{\log 36}{\log 8}\right)}+\dfrac{1}{\left(\dfrac{\log 36}{\log 9}\right)}+\dfrac{1}{\left(\dfrac{\log 36}{\log 18}\right)}\end{aligned}$

$\Longrightarrow\ \ \dfrac{\log 8}{\log 36}+\dfrac{\log 9}{\log 36}+\dfrac{\log 18}{\log 36}\\ \\ \\ \Longrightarrow\ \ \dfrac{\log 8+\log 9+\log 18}{\log 36}\\ \\ \\ \Longrightarrow\ \ \dfrac{\log(8\cdot 9\cdot 18)}{\log 36}\\ \\ \\ \Longrightarrow\ \ \dfrac{\log(2^3\cdot 3^2\cdot 3^2\cdot 2)}{\log(6^2)}\\ \\ \\ \Longrightarrow\ \ \dfrac{\log(2^4\cdot 3^4)}{2\log 6}\\ \\ \\ \Longrightarrow\ \ \dfrac{\log((2\cdot 3)^4)}{2\log(2\cdot 3)}$

$\\ \\ \\ \Longrightarrow\ \ \dfrac{4\log(2\cdot 3)}{2\log(2\cdot 3)}\\ \\ \\ \Longrightarrow\ \ \dfrac{4}{2}\\ \\ \\ \Longrightarrow\ \ 2\\ \\ \\ \Longrightarrow\ \ \textsf{RHS}$

Hence Proved!