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First thing's first - do you know what the domain of log (a) is ?

a>0

From that we can deduce that log (x^2-5x+16) has to be <1 so that 1 - log (x^2-5x+16) >0

ok

But at the same time, for log (x^2-5x+16) to exist, (x^2-5x+16) also has to be >0.

And to my guess, the quadratic itself will look something like this
|dw:1435347000458:dw|

is the answer
\(-3

. . .

-3

Yeah, it's the same thing really, solve either one of them and you'll get there.

\(\Large \left\{ \begin{gathered}
\frac{{10}}{{{x^2} - 5x + 16}} > 1 \hfill \\
\hfill \\
{x^2} - 5x + 16 > 0 \hfill \\
\end{gathered} \right.\)
by evaluating this i got
\(2

Simply taking one example from your answer.
If 2

r u sure that \(10/9.75\) is not greater than \(1\)

Oh wait, damn.

Ahahah.

Yeah, it's 2

Big medal for whoever finds the point at which I went wrong with my asumptions.

Oh, got it.

everything is correct except u graphed wrong

The roots of the quadratic were 2 and 3 not -2 and -3.

Big blunder on my side, I apologize.

np thnks