Question

Hii frnds plz I need the answer for this question​

Answer 1

1 is the answer HOPE IT WILL HELP YOU

Answer 2

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Some notes :

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

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

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