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

Question

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 (A) e 1 / 5 (B) √ e  (C) √[3] e  (D) √[4] e

Tardigrade Admin · Accepted Answer

( ln a 2/1- ln a 2)=(3 ln a /1- ln a ) arrow (2 ln a /1-2 ln a )=(3 ln a /1- ln a ) arrow 2( ln a )-2( ln a )2=3 ln a -6( ln a )2 arrow 4( ln a )2- ln a =0 arrow ln a =0 or (1/4). Thus, .a =1, e 1 / 4]

Q. If $a \neq = 1$ and $ln a^{2} + (ln a^{2})^{2} + (ln a^{2})^{3} + \dots\dots .. = 3 (ln a + (ln a)^{2} + (ln a)^{3} + (ln a)^{4} + \dots\dots ..)$ then 'a' is equal to

$e^{1/5}$

$e$

$3 e$

$4 e$

$\frac{l n a ^{2}}{1 - l n a ^{2}} = \frac{3 l n a}{1 - l n a} \to \frac{2 l n a}{1 - 2 l n a} = \frac{3 l n a}{1 - l n a} \to 2 (ln a) - 2 (ln a)^{2} = 3 ln a - 6 (ln a)^{2}$ $\to 4 (ln a)^{2} - ln a = 0 \to ln a = 0$ or $\frac{1}{4}$ . Thus, $a = 1, e^{1/4}]$

Q. If a=1 and lna2+(lna2)2+(lna2)3+……..=3(lna+(lna)2+(lna)3+(lna)4+……..) then 'a' is equal to

e1/5

e​

3e​

4e​

1−lna2lna2​=1−lna3lna​→1−2lna2lna​=1−lna3lna​→2(lna)−2(lna)2=3lna−6(lna)2 →4(lna)2−lna=0→lna=0 or 41​. Thus, a=1,e1/4]

Q. If $a \neq = 1$ and $ln a^{2} + (ln a^{2})^{2} + (ln a^{2})^{3} + \dots\dots .. = 3 (ln a + (ln a)^{2} + (ln a)^{3} + (ln a)^{4} + \dots\dots ..)$ then 'a' is equal to

$e^{1/5}$

$e$

$3 e$

$4 e$

$\frac{l n a ^{2}}{1 - l n a ^{2}} = \frac{3 l n a}{1 - l n a} \to \frac{2 l n a}{1 - 2 l n a} = \frac{3 l n a}{1 - l n a} \to 2 (ln a) - 2 (ln a)^{2} = 3 ln a - 6 (ln a)^{2}$ $\to 4 (ln a)^{2} - ln a = 0 \to ln a = 0$ or $\frac{1}{4}$ . Thus, $a = 1, e^{1/4}]$