Question

ans for ques 65 pls help me.....

Answer 1

Answer:

d)  x = a^{1/a}

Hello.  I hope this helps you.  Have a great day!!!

Step-by-step explanation:

Hmmm.  The typesetting looks pretty bad.

The bases of those logs, I am certain, are not meant to be

$3\sqrt{a},\dots,a\sqrt{a}$

but rather

$\sqrt[3]{a},\dots,\sqrt[a]{a}$

I will work on that assumption.

For convenience, all logs here without a base written in are logs "base a".

Need to know that:

$\displaystyle\log_c b = \frac{\log b}{\log c}$

So

$\displaystyle\log_a x + \log_{\sqrt a}x + \log_{\sqrt[3]a}x +\cdots+\log_{\sqrt[a]a}x\\ \\=\log x + \frac{\log x}{\log\sqrt a} + \frac{\log x}{\log\sqrt[3]a} +\cdots+\frac{\log x}{\log\sqrt[a]a}\\ \\=\log x + \frac{\log x}{\frac12\log a} + \frac{\log x}{\frac13\log a} +\cdots+\frac{\log x}{\frac1a\log a}\\ \\=\frac{\log x}{\log a}(1 + 2 + 3 +\cdots+a)\\ \\=\frac{\log x}{\log a}\,\frac{a(a+1)}{2} = \frac{a+1}{2}\\ \\\Rightarrow \log x = \tfrac1a\log a = \log a^{1/a}\\ \\\Rightarrow x = a^{1/a}$