Question

Solve this question of sequence and series

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: 1, \: log_{9}( {3}^{1 - x} + 2), \: log_{3}(4. {3}^{x} - 1) \: are \: in \: AP$

We know, If a, b, c are in AP, then 2b = a + c

So, using this identity, we get

$\rm \: 2log_{9}( {3}^{1 - x} + 2) \: = \:1 + log_{3}(4. {3}^{x} - 1)$

$\rm \: 2log_{ {3}^{2} }( {3}^{1 - x} + 2) \: = \: log_{3}(3) + log_{3}(4. {3}^{x} - 1)$

We know,

$\boxed{ \rm{ \: log_{ {a}^{x} }( {b}^{y} ) = \frac{y}{x} log_{a}(b) \: }}$

and

$\boxed{ \rm{ \:logx + logy = log(xy) \: }}$

So, using these results, we get

$\rm \: 2 \times \frac{1}{2} log_{3}( {3}^{1 - x} + 2) \: = \: log_{3}3[4. {3}^{x} - 1]$

$\rm \: log_{3}( {3}^{1 - x} + 2) \: = \: log_{3}[12. {3}^{x} - 3]$

$\rm \: {3}^{1 - x} + 2 \: = \: 12. {3}^{x} - 3$

can be further rewritten as

$\rm \:\dfrac{3}{ {3}^{x} } \: = \: 12. {3}^{x} - 5$

$\rm \: \: 12. ({3}^{x})^{2} - 5.{3}^{x} = 3$

$\rm \: \: 12. ({3}^{x})^{2} - 5.{3}^{x} - 3 = 0$

$\rm \: \: 12. ({3}^{x})^{2} - 9.{3}^{x} + 4.{3}^{x} - 3 = 0$

$\rm \: \: 3. {3}^{x}(4.{3}^{x} - 3) + (4.{3}^{x} - 3) = 0$

$\rm \: \: (4.{3}^{x} - 3)(3.{3}^{x} + 1) = 0$

$\rm\implies \:{3}^{x} = \dfrac{3}{4} \: \: or \: \: {3}^{x} = - \: \dfrac{1}{3} \: \{rejected \: as \: {3}^{x} > 0 \}$

$\rm \: {3}^{x} = \dfrac{3}{4}$

Taking log on both sides, we get

$\rm \:log{3}^{x} = log\bigg(\dfrac{3}{4}\bigg)$

$\rm \: xlog3 = log3 - log4$

$\rm \: x = \dfrac{log3}{log3} - \dfrac{log4}{log3}$

$\bf\implies \:x = 1 - log_{3}(4)$

Hence, Option (2) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

\large\underline{\sf{Solution-}}

Solution−

Given that,

\begin{gathered}\rm \: 1, \: log_{9}( {3}^{1 - x} + 2), \: log_{3}(4. {3}^{x} - 1) \: are \: in \: AP \\ \end{gathered}

1,log

9

(3

1−x

+2),log

3

(4.3

x

−1)areinAP

We know, If a, b, c are in AP, then 2b = a + c

So, using this identity, we get

\begin{gathered}\rm \: 2log_{9}( {3}^{1 - x} + 2) \: = \:1 + log_{3}(4. {3}^{x} - 1) \\ \end{gathered}

2log

9

(3

1−x

+2)=1+log

3

(4.3

x

−1)

\begin{gathered}\rm \: 2log_{ {3}^{2} }( {3}^{1 - x} + 2) \: = \: log_{3}(3) + log_{3}(4. {3}^{x} - 1) \\ \end{gathered}

2log

3

2

(3

1−x

+2)=log

3

(3)+log

3

(4.3

x

−1)

We know,

\begin{gathered}\boxed{ \rm{ \: log_{ {a}^{x} }( {b}^{y} ) = \frac{y}{x} log_{a}(b) \: }} \\ \end{gathered}

log

a

x

(b

y

)=

x

y

log

a

(b)

and

\begin{gathered}\boxed{ \rm{ \:logx + logy = log(xy) \: }} \\ \end{gathered}

logx+logy=log(xy)

So, using these results, we get

\begin{gathered}\rm \: 2 \times \frac{1}{2} log_{3}( {3}^{1 - x} + 2) \: = \: log_{3}3[4. {3}^{x} - 1] \\ \end{gathered}

2×

2

1

log

3

(3

1−x

+2)=log

3

3[4.3

x

−1]

\begin{gathered}\rm \: log_{3}( {3}^{1 - x} + 2) \: = \: log_{3}[12. {3}^{x} - 3] \\ \end{gathered}

log

3

(3

1−x

+2)=log

3

[12.3

x

−3]

\begin{gathered}\rm \: {3}^{1 - x} + 2 \: = \: 12. {3}^{x} - 3\\ \end{gathered}

3

1−x

+2=12.3

x

−3

can be further rewritten as

\begin{gathered}\rm \:\dfrac{3}{ {3}^{x} } \: = \: 12. {3}^{x} - 5\\ \end{gathered}

3

x

3

=12.3

x

−5

\begin{gathered}\rm \: \: 12. ({3}^{x})^{2} - 5.{3}^{x} = 3\\ \end{gathered}

12.(3

x

)

2

−5.3

x

=3

\begin{gathered}\rm \: \: 12. ({3}^{x})^{2} - 5.{3}^{x} - 3 = 0\\ \end{gathered}

12.(3

x

)

2

−5.3

x

−3=0

\begin{gathered}\rm \: \: 12. ({3}^{x})^{2} - 9.{3}^{x} + 4.{3}^{x} - 3 = 0\\ \end{gathered}

12.(3

x

)

2

−9.3

x

+4.3

x

−3=0

\begin{gathered}\rm \: \: 3. {3}^{x}(4.{3}^{x} - 3) + (4.{3}^{x} - 3) = 0\\ \end{gathered}

3.3

x

(4.3

x

−3)+(4.3

x

−3)=0

\begin{gathered}\rm \: \: (4.{3}^{x} - 3)(3.{3}^{x} + 1) = 0\\ \end{gathered}

(4.3

x

−3)(3.3

x

+1)=0

\begin{gathered}\rm\implies \:{3}^{x} = \dfrac{3}{4} \: \: or \: \: {3}^{x} = - \: \dfrac{1}{3} \: \{rejected \: as \: {3}^{x} > 0 \} \\ \end{gathered}

⟹3

x

=

4

3

or3

x

=−

3

1

{rejectedas3

x

>0}

\begin{gathered}\rm \: {3}^{x} = \dfrac{3}{4} \\ \end{gathered}

3

x

=

4

3

Taking log on both sides, we get

\begin{gathered}\rm \:log{3}^{x} = log\bigg(\dfrac{3}{4}\bigg) \\ \end{gathered}

log3

x

=log(

4

3

)

\begin{gathered}\rm \: xlog3 = log3 - log4 \\ \end{gathered}

xlog3=log3−log4

\begin{gathered}\rm \: x = \dfrac{log3}{log3} - \dfrac{log4}{log3} \\ \end{gathered}

x=

log3

log3

−

log3

log4

\begin{gathered}\bf\implies \:x = 1 - log_{3}(4) \\ \end{gathered}

⟹x=1−log

3

(4)

Hence, Option (2) is correct.

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★log

x

(x)=1

★log

x

(x

y

)=y

★log

x

z

(x

w

)=

z

w

★log

a

(b)=

loga

logb

★e

logx

=x

★e

ylogx

=x

y

★log1=0