Math, asked by aa3044954, 19 days ago


 log_{2}\pi  +  log_{4}\pi +  log_{8}\pi =  \frac{22}{6} then \: \pi =
1)1
2)2
3)3
4)4 ​

Answers

Answered by mathdude500
5

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

Given logarithmic equation is

\rm \: log_{2}\pi + log_{4}\pi + log_{8}\pi = \dfrac{22}{6}

can be rewritten as

\rm \: log_{2}\pi + log_{ {2}^{2} }\pi + log_{ {2}^{3} }\pi = \dfrac{11}{3}

We know,

\boxed{\rm{  \: log_{ {x}^{y} }( {x}^{z} )  =  \frac{z}{y} \: }} \\

So, using this result, we get

\rm \: log_{2}\pi +  \dfrac{1}{2} log_{2}\pi + \dfrac{1}{3}  log_{2}\pi = \dfrac{11}{3} \\

\rm \: log_{2}\pi \bigg(1+  \dfrac{1}{2}  + \dfrac{1}{3}\bigg) = \dfrac{11}{3} \\

\rm \: log_{2}\pi \bigg(\dfrac{6 + 3 + 2}{6}\bigg) = \dfrac{11}{3} \\

\rm \: log_{2}\pi \bigg(\dfrac{11}{6}\bigg) = \dfrac{11}{3} \\

\rm \: log_{2}\pi = 2 \\

We know,

\boxed{\rm{  \: log_{a}(b) = c\rm\implies \:b =  {a}^{c} \: }} \\

So, using this result, we get

\rm \: \pi \:  =  \:  {2}^{2}  \\

\rm\implies \:\rm \: \pi \:  =  \:  4 \\

Hence,

\boxed{\rm{  \:log_{2}\pi + log_{4}\pi + log_{8}\pi = \frac{22}{6}  \: then \: \pi =4 \: }} \\

So, option (4) 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{logxy = logx + logy}\\ \\ \bigstar \: \bf{log \dfrac{x}{y} = logx - logy }\\ \\ \bigstar \: \bf{ log {x}^{y} = y \: log}\\ \\ \bigstar \: \bf{ log_{x}(x)  = 1}\\ \\ \bigstar \: \bf{ log_{ {x}^{y} }(x)  =  \dfrac{1}{y} }\\ \\ \bigstar \: \bf{ {a}^{ log_{a}(x) }  = x}\\ \\ \bigstar \: \bf{ {a}^{y \:  log_{a}(x) }  =  {x}^{y} }\\ \\ \bigstar \: \bf{ {e}^{logx}  = x}\\ \\ \bigstar \: \bf{ {e}^{y \: logx}  \:  =  \:  {x}^{y} }\\ \\  \end{array} }}\end{gathered}\end{gathered}\end{gathered}

Answered by Anonymous
5

 log_{2}\pi +  log_{4}\pi +  log_{8}\pi =  \frac{22}{6}    \\  log_{2}\pi +  log_{ {2}^{2} }\pi +  log_{ {2}^{3} }\pi =  \frac{22}{6}

⟶ we know that ( log_{ {b}^{n} }a =  \frac{1}{n} log_{b}a)

 log_{2}\pi + \frac{1}{2}   log_{2}\pi +   \frac{1}{3}  log_{2}\pi  =  \frac{22}{6}  \\  \frac{11}{6} log_{2}\pi =  \frac{22}{6}    \\  log_{2}\pi = 2

⟹ π = 4

Similar questions