The value of x satisfying log_a x + log_{sqrt | vedclass

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(A) The given equation is: $\log_a x + \log_{a^{1/2}} x + \log_{a^{1/3}} x + \dots + \log_{a^{1/a}} x = \frac{a(a+1)}{2}$.Using the property $\log_{a^

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$x = a$

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(A) The given equation is: $\log_a x + \log_{a^{1/2}} x + \log_{a^{1/3}} x + \dots + \log_{a^{1/a}} x = \frac{a(a+1)}{2}$.Using the property $\log_{a^k} x = \frac{1}{k} \log_a x$,we can rewrite the terms:$\log_a x + 2 \log_a x + 3 \log_a x + \dots + a \log_a x = \frac{a(a+1)}{2}$.Factor out $\log_a x$:$\log_a x (1 + 2 + 3 + \dots + a) = \frac{a(a+1)}{2}$.The sum of the first $a$ natural numbers is $\frac{a(a+1)}{2}$.So,$\log_a x \cdot \frac{a(a+1)}{2} = \frac{a(a+1)}{2}$.Dividing both sides by $\frac{a(a+1)}{2}$,we get $\log_a x = 1$.Therefore,$x = a^1 = a$.

The value of $x$ satisfying $\log_a x + \log_{\sqrt{a}} x + \log_{\sqrt[3]{a}} x + \dots + \log_{\sqrt[a]{a}} x = \frac{a(a+1)}{2}$ is:

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