Logarithm Solution Doubt

Asked Mar 15 '22 at 09:50

Active Mar 15 '22 at 09:50

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I am getting two real solutions to this question, but the answer key says there is only one. Could anyone please say why one solution got rejected?

$$9^{\log _3\left(\log _e\left(x\right)\right)}=\log _e\left(x\right)-\left(\log \:_e\left(x\right)\right)^2+1$$

I got two solutions $\log _e\left(x\right)=-\frac{1}{2}$ and $\log _e\left(x\right)=1$.

asked Mar 15 '22 at 09:50

The negative solution is impossible, as you can't have $\log_3 \left(-\frac 12\right)$. – lulu Mar 15 '22 at 10:06
Yeah, this logarithm is impossible. But $\log _e\left(x\right)=-\frac{1}{2}$ and $\log _3\left(-\frac{1}{2}\right)$ are different. The answer of $\log _e\left(x\right)=-\frac{1}{2}$ gives you $x=\frac{1}{\sqrt{e}}$. – Anonymousstriker38596 Mar 15 '22 at 13:03
Oh wait, nvm I got what you were trying to say. – Anonymousstriker38596 Mar 15 '22 at 13:04

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