Find $x$ in $\log x^2 = (\log x)^2$

Question

Find $x$ in $\log x^2 = (\log x)^2$.

I couldn't find x.

They are supposed to italisize – user3175999 Jan 29 '14 at 19:08 — user3175999, Jan 29 '14 at 19:08
Hint: use $\log(x^2) = 2\log x$ for positive $x$. – GEdgar Jan 29 '14 at 19:13 — GEdgar, Jan 29 '14 at 19:13

score 4 · Answer 1 · answered Jan 29 '14 at 19:15

4

$$\log x^2 = (\log x)^2 \\2\log x = (\log x)^2 \\ \log x(\log x-2)=0 \\ \log x=0 \to x=1 \\ \log x=2 \to x=10^2$$

answered Jan 29 '14 at 19:15

Semsem

thanks, but the other person's way was a bit more clear. – user3175999 Jan 29 '14 at 19:51
Never mind, as always if you face any problem just let me know – Semsem Jan 29 '14 at 19:53

score 2 · Accepted Answer · answered Jan 29 '14 at 19:39

2

Hint: Note that $$\log x^2=2\log x$$ for all $x>0.$ Now use the substitution $u=\log x$ to get $$2u=u^2.$$ Solve for $u,$ then solve for $x.$

answered Jan 29 '14 at 19:39

Cameron Buie

thanks, your way helped me understand it a bit better. – user3175999 Jan 29 '14 at 19:50
You are very welcome! – Cameron Buie Jan 29 '14 at 21:20

2 Answers2