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So after trying to solve the following inequality, log(x) + log(2-x) < 1 I couldn't end up with any answer since as I tried to solve here are my results

First I got my reference equation ( to be used in the number line ) log(x)+log(2-x)<1

As I tried solving it

x(2-x)<10

x(2-x)-10<0

x^2 + 2x -10 <0

I multiply both sides by negative 1

x^2 - 2x + 10 > 0

But as you can see the following is not factorable.

I decided to try using Symbolab and Wolfram to see what was the Interval notation and it answered (0,2)

Can someone please help enlighten me or atleast point me in the right way to solve the following inequality

Many thanks!!

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$\log(x)$ is defined for $x>0$

and $\log(2-x)$ is for $2-x>0\iff x<2$

So, we need $0<x<2$

Now $x(2-x)<10\iff x^2-2x+10>0$

Now for real $x,x^2-2x+10=(x-1)^2+9\ge9>0$

Can you take it from here?

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  • $\begingroup$ Yes! Thanks I understood it now! $\endgroup$ – Sam P Aug 6 '16 at 11:49
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For $x\in(0,2)$ we have:

$$\log (x)+\log(2-x)=\log[x(2-x)].$$

Since $x(2-x)$ has roots of $0$ and $2$, its maximum is at $1$. Thus $x(2-x)\leq 1$. Therefore $$\log(x)+\log(2-x)\leq \log(1)= 0<1$$ for all $x\in(0,2)$.

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