Question

If $a\ne 1$ and $ln{a}^2+{\left(ln{a}^2\right)}^2+{\left(ln{a}^2\right)}^3+........=3(lna+{\left(lna\right)}^2+{\left(lna\right)}^3+{\left(lna\right)}^4+......)$ then $a$ is

• A. $anirrationalnumber$
• B. $atranscendentalnumber$
• C. $analgeberaicnumber$
• D. $asurd$

Answer 1

The correct option is A $anirrationalnumber$

Given, for $a\ne 1$,

$ln{a}^2+{\left(ln{a}^2\right)}^2+{\left(ln{a}^2\right)}^3+........=3(lna+{\left(lna\right)}^2+{\left(lna\right)}^3+{\left(lna\right)}^4+......)$

or, $\frac{\ln a^2}{1-\ln a^2}=3\times \frac{\ln a}{1-\ln a}$

or, $\frac{2\ln a}{1-2\ln a}=3\times \frac{\ln a}{1-\ln a}$

or, $2(1-\ln a)=3(1-2\ln a)$ [Since $a\ne 1\Rightarrow \ln a\ne 0$ ]

or, $4\ln a=1$

or, $a=\sqrt[4]{4e}$.

So clearly $a$ is an irrational number.