1. Originally Posted by SVXX
Ok, the last problem.

$|\log_{2}\left(\frac{4-x}{2x - 5}\right)| + \log_{2}x = 1$

....
Here is another attempt to solve the equation:

1. On behalve of the summand $\log_{2}x$ x must be greater than zero.

2.
$\left| \log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x = 1$

$\left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| = \log_{2}\left(\frac2x\right)$

You have to consider 2 different cases:

$\log_{2}\left(\frac{4-x}{2x - 5}\right) > 0~\implies~ \frac{4-x}{2x - 5} > 1$

Then you must solve for x:

$\frac{4-x}{2x - 5} = \frac2x~\implies~ \boxed{x = \sqrt{10}}$

The negative solution isn't valid. See #1.

Or

$\log_{2}\left(\frac{4-x}{2x - 5}\right) < 0~\implies~ - \log_{2}\left(\frac{4-x}{2x - 5}\right) > 0 ~\implies~ \frac{2x - 5}{4-x} > 1$

Then you have to solve for x:

$\frac{2x - 5}{4-x} = \frac2x ~\implies~ \boxed{x \approx 2.886}$

The negative solution isn't valid. See #1.

Edit: To my great confusion both solutions don't satisfy the constraints to yield a fraction which is greater than 1.

2. Originally Posted by SVXX
In that case, x = -1.38!
I forgot something, I remembered that when taking my shower .. .This one doesn't work either because we have $\log x$, meaning that x has to be > 0...

3. Originally Posted by SVXX
Ok, the last problem.

$|\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1$
...
To illustrate that your equation don't have real solutions I've sketched the graphs of

$f(x)=\left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x$

and

$g(x)=1$

The graphs don't intersect. That means there doesn't exist an x so that the values of the functions are equal.

4. Originally Posted by earboth
To illustrate that your equation don't have real solutions I've sketched the graphs of

$f(x)=\left|\log_{2}\left(\frac{4-x}{2x - 5}\right)\right| + \log_{2}x$

and

$g(x)=1$

The graphs don't intersect. That means there doesn't exist an x so that the values of the functions are equal.
You're right, I went through the properties again..x cannot possibly be negative. That means these equations have no solution as per you peeps.
That'll be interesting. Wonder what the teacher will say when I point that out .
Anyways thanks for pointing out this huge blunder!
I can write x doesn't belong to R then I guess.

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