1. Tardigrade
  2. Question
  3. Mathematics
  4. If a ≠ 1 and ln a 2+( ln a 2)2+( ln a 2)3+ ldots ldots . .=3( ln a +( ln a )2+( ln a )3+( ln a )4+ ldots ldots ..) then 'a' is equal to

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Q. If $a \neq 1$ and $\ln a ^2+\left(\ln a ^2\right)^2+\left(\ln a ^2\right)^3+\ldots \ldots . .=3\left(\ln a +(\ln a )^2+(\ln a )^3+(\ln a )^4+\ldots \ldots ..\right)$ then 'a' is equal to

Sequences and Series

Solution:

$\quad \frac{\ln a ^2}{1-\ln a ^2}=\frac{3 \ln a }{1-\ln a } \rightarrow \frac{2 \ln a }{1-2 \ln a }=\frac{3 \ln a }{1-\ln a } \rightarrow 2(\ln a )-2(\ln a )^2=3 \ln a -6(\ln a )^2$ $\rightarrow 4(\ln a )^2-\ln a =0 \rightarrow \ln a =0$ or $\frac{1}{4}$. Thus, $\left.a =1, e ^{1 / 4}\right]$