Math, asked by rahularnipalli, 30 days ago

Hii
plz tell me the answer of this question​

Attachments:

Answers

Answered by guptavirag002
1

Answer:

Check it out ! Hope it will help u !

Attachments:
Answered by TrustedAnswerer19
2

\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}

Some notes :

1 + 2 + 3 + 4 +  . \: . \: . + n \\  =  \frac{n(n + 1)}{2}  \\  \\  log_{x}x = 1  \:  \:  \:  \therefore \:  log_{2}2 = 1 \\  \\  log_{a} {x}^{n}  = n \times  log_{a}x \\  \\ now \\

 log_{x}2 +  log_{x} {2}^{2}  +  log_{x} {2}^{3}  + . \: . \: . \:  +  log_{x} {2}^{n}  \\  = log_{x}2 +  2 \: log_{x} {2}  + \: 3   \: log_{x} {2} + . \: . \: . \:  + n \:  log_{x} {2} \\  = (1 + 2 + 3 + ... + n) log_{x}2 \\  =  \frac{n(n + 1)}{2}  \times  log_{x}2 \\  \\ if \: x = 2 \:  \: then \:  log_{2}2 = 1 \\  =  (1 + 2 + 3 + ... + n) log_{x}2 \\  =  \frac{n(n + 1)}{2}  \times  log_{2}2  \\  =  (1 + 2 + 3 + ... + n) log_{x}2 \\  =  \frac{n(n + 1)}{2}   \:  \:  \therefore \: x = 2

Attachments:
Similar questions