Question

logx ^2+logx2^2+logx2^3........... +logx2^n=n(n+1)/2 then x=​

Answer 1

Step-by-step explanation:

$\sf {log}_{x}{2} + {log}_{x}{{2}^{2}} + {log}_{x}{{2}^{3}} + {log}_{x}{{2}^{4}} + ... + {log}_{x}{{2}^{n}} = \frac{n(n+1)}{2} \\ \\ \sf \longrightarrow {log}_{x}{2} + 2 {log}_{x}{2} + 3{log}_{x}{2} + ... + n{log}_{x}{2} = \frac{n(n+1)}{2} \\ \\ \sf \longrightarrow {log}_{x}{2}( 1 + 2 + 3 + ... + n ) = \frac{n(n+1)}{2} \\ \\ \sf \longrightarrow {log}_{x}{2}\frac{n(n+1)}{2} = \frac{n(n+1)}{2} \\ \\ \sf \longrightarrow {log}_{x}{2} = 1 \\ \\ \sf \longrightarrow {x}^{1} = 2 \\ \\ \sf \longrightarrow x = 2 \\ \\ \sf EXTRA\: INFORMATION \\ \\ \sf \star \star {log}_{x}{{a}^{m}} = m{log}_{x}{a} \\ \\ \sf \star \star {log}_{x}{a} + {log}_{x}{b} = {log}_{x}{ab} \\ \\ \sf \star \star {log}_{x}{a} - {log}_{x}{b} = {log}_{x}{\frac{a}{b}} \\ \\ \sf \star \star \: Sum\:of\:n\: natural\:number = \frac{n(n+1)}{2}$

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