${\log _2}.{\log _3}....{\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{{.2}^1}}}}}}}}}}}= . . . $.
Medium
Download our app for free and get started
b
(b) \({\log _2}.{\log _3}.....{\log _{99}}\) \({\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{.^{{2^1}}}}}}}}}}}}}\)
(b) \({\log _2}.{\log _3}.....{\log _{99}}\) \({\log _{100}}{100^{{{99}^{{{98}^{{.^{{.^{{.^{{2^1}}}}}}}}}}}}}\)
\( = {\log _2}.{\log _3}....{\log _{98}}^{{{98}^{{{97}^{{.^{{.^{{.^{{2^1}}}}}}}}}}}}\)
\( = {\log _2}\,\,2'{\log _3}3 = {\log _2}2 = 1\).
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1જો ${x \over {(x - 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x - 1)}} - {{x + 1} \over {{x^2} + 1}}} \right] + y$ તો $y =$View Solution
- 2${(0.05)^{{{\log }_{_{\sqrt {20} }}}(0.1 + 0.01 + 0.001 + ......)}}= . .$ . .View Solution
- 3${{{x^4} + 24{x^2} + 28} \over {{{({x^2} + 1)}^3}}} = . .$ .View Solution
- 4સમીકરણ $log_7(2^x -1) + log_7(2^x -7) = 1$ ના ઉકેલોની સંખ્યા મેળવો.View Solution
- 5જો ${a^x} = {b^y} = {(ab)^{xy}},$ તો $x + y = $View Solution
- 6$\sqrt {[12 - \sqrt {(68 + 48\sqrt 2 )} ]} = $View Solution
- 7જો ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0$ તો $x = . . . .$View Solution
- 8${{{x^2} + 1} \over {(2x - 1)\,({x^2} - 1)}} = $View Solution
- 9જો $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$ તો આપેલ પૈકી કોની કિમત $1$ છે.View Solution
- 10જો ${1 \over {x(x + 1)\,(x + 2)....(x + n)}} = {{{A_0}} \over x} + {{{A_1}} \over {x + 1}} + {{{A_2}} \over {x + 2}} + .... + {{{A_n}} \over {x + n}}$ તો ${A_r} = $View Solution