Answers
Solution:
\underline{\textsf{To Find : }}
To Find :
\sf = \: \dfrac{1}{log_3 \: e } \: + \: \dfrac{1}{log_3 \: e^2} \: + \: \dfrac{1}{log_3 \: e^4} \: + \: .....=
log
3
e
1
+
log
3
e
2
1
+
log
3
e
4
1
+.....
\begin{lgathered}\textsf{Using Logarithm Rule : } \\ \\ \boxed{\underline{\mathsf{ \implies log_{a} \: b^c \: = \: c \: log_a \: b }}}\end{lgathered}
Using Logarithm Rule :
⟹log
a
b
c
=clog
a
b
\sf = \: \dfrac{1}{log_3 \: e } \: + \: \dfrac{1}{2 \: log_3 \: e} \: + \: \dfrac{1}{4 \: log_3 \: e} \: + \: .....=
log
3
e
1
+
2log
3
e
1
+
4log
3
e
1
+.....
\sf Take \: out \: \: \dfrac{1}{log_3 \: e} \: \: as \: common \: factor .Takeout
log
3
e
1
ascommonfactor.
\begin{lgathered}\sf = \: \dfrac{1}{log_3 \: e } \underbrace{\Bigg( 1 \: + \: \dfrac{1}{2} \: + \: \dfrac{1}{4} \: + ..... \Bigg) } \\ \textsf{ \: \: \: \: \: \: \: This Part forms an infinte G.P.}\end{lgathered}
=
log
3
e
1
(1+
2
1
+
4
1
+.....)
This Part forms an infinte G.P.
\begin{lgathered}\textsf{Where, } \\ \\ \mathsf{\implies First \: term (a) \: = \: 1 } \\ \\ \\ \mathsf{\implies Common \: ratio(r) \: = \: \dfrac{\dfrac{1}{2}}{ \: \: \: \: 1 \: \: \: \: } \: = \: \dfrac{1}{2} }\end{lgathered}
Where,
⟹Firstterm(a)=1
⟹Commonratio(r)=
1
2
1
=
2
1
\begin{lgathered}\underline{\textsf{Using Formula :}} \\ \\ \mathsf{\implies Sum \: of \: infinite \: G.P. \: = \: \dfrac{a}{1 \: - \: r}}\end{lgathered}
Using Formula :
⟹SumofinfiniteG.P.=
1−r
a
\begin{lgathered}\sf = \: \dfrac{1}{log_3 \: e} \left( \dfrac{1}{1 \: - \: \dfrac{1}{2}} \right) \\ \\ \\ \sf = \: \dfrac{1}{log_3 \: e} \left( \dfrac{1}{\dfrac{2 \: - \: 1}{2}} \right)\end{lgathered}
=
log
3
e
1
⎝
⎜
⎛
1−
2
1
1
⎠
⎟
⎞
=
log
3
e
1
⎝
⎜
⎛
2
2−1
1
⎠
⎟
⎞
\sf = \: \dfrac{1}{log_3 \: e} \left( \dfrac{ \: \: \: \: 1 \: \: \: \: }{\dfrac{1}{2}} \right)=
log
3
e
1
⎝
⎜
⎛
2
1
1
⎠
⎟
⎞
\begin{lgathered}\sf = \: \dfrac{1}{log_3 \: e}\left( 1 \: \div \: \dfrac{1}{2} \right) \\ \\ \\ \sf = \: \dfrac{1}{log_3 \: e}\left( 1 \: \times \: 2\right)\end{lgathered}
=
log
3
e
1
(1÷
2
1
)
=
log
3
e
1
(1×2)
\begin{lgathered}\sf = \: \dfrac{2}{log_3 \: e} \\ \\ \\ \sf = \: \dfrac{2 \: (log_3 \: 3)}{log_3 \: e} \qquad \quad \boxed{ \mathsf{\implies log_3 \: 3 \: = \: 1} }\end{lgathered}
=
log
3
e
2
=
log
3
e
2(log
3
3)
⟹log
3
3=1
\begin{lgathered}\sf = \: \dfrac{ \: \: \: log_3 \: 3^2 \: \: \: }{log_3 \: e} \\ \\ \\ \sf = \: \dfrac{ \: \: \: log_3 \: 9 \: \: \: }{log_3 \: e}\end{lgathered}
=
log
3
e
log
3
3
2
=
log
3
e
log
3
9
\begin{lgathered}\textsf{Using Logarithm Rule : } \\ \\ \boxed{\underline{\mathsf{ \implies \dfrac{ \: \: \: log_{a} \: b \: \: \: }{ log_{a} \: c } \: = \: log_{c} \: b }}}\end{lgathered}
Using Logarithm Rule :
⟹
log
a
c
log
a
b
=log
c
b
\sf = \: log_e \: 9=log
e
9
\underline{\underline{\textsf{The correct option is (A). }}}
The correct option is (A).