\(\large \color{black}{\begin{align}& \normalsize \text{Find the domain of the function} \hspace{.33em}\\~\\
& f(x)=\log_{10} [1-\log_{10} (x^2-5x+16)] \hspace{.33em}\\~\\
\end{align}}\)

- mathmath333

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- anonymous

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

- mathmath333

a>0

- anonymous

Precisely! So from that we can deduce so much that 1 - log (x^2-5x+16) has to be >0 for that whole thing to exist.

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## More answers

- anonymous

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

- mathmath333

ok

- anonymous

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

- Michele_Laino

we have this condition, for the existence of the outer logarithm:
\[\Large \begin{gathered}
1 - {\log _{10}}\left( {{x^2} - 5x + 16} \right) > 0 \hfill \\
\hfill \\
{\log _{10}}\left[ {\frac{{10}}{{{x^2} - 5x + 16}}} \right] > 0 \hfill \\
\hfill \\
\frac{{10}}{{{x^2} - 5x + 16}} > 1 \hfill \\
\end{gathered} \]

- Michele_Laino

furthermore we have the subsequent condition for the existence of the inner logarithm:
\[\Large {x^2} - 5x + 16 > 0\]

- Michele_Laino

then the domain of our function is given by the solution of this system:
\[\Large \left\{ \begin{gathered}
\frac{{10}}{{{x^2} - 5x + 16}} > 1 \hfill \\
\hfill \\
{x^2} - 5x + 16 > 0 \hfill \\
\end{gathered} \right.\]

- anonymous

It sounds a little wrapped up but it will make sense along the way.
Let's work this one step at a time.
For log (x^2-5x+16) to exist, (x^2-5x+16) also has to be >0.
We solve the quadratic we find that (x^2-5x+16) has complex roots and that means that (x^2-5x+16) is always positive so we need not worry about it!
Now, for that whole thing to exist, we've concluded that 1 - log (x^2-5x+16) >0
We "flip" the log over a little and we have that
1 > log (x^2-5x+16)
Express 1 as log (10) ( log(10) is 1 ) and you have
log (10) > log (x^2-5x+16)
Which equates to
10 > (x^2-5x+16)
Or
0 > (x^2-5x+6)
We solve this new quadratic and we have that (x^2-5x+6)=(x-2)*(x-3).
What that means is that (x^2-5x+6) will look something like this:
|dw:1435346669499:dw|

- anonymous

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

- anonymous

Therefore, for 0>x^2-5x+6, the quadratic must be negative and that only happens for x between (-3,-2).

- mathmath333

is the answer
\(-3

- anonymous

. . .

- anonymous

-3

- anonymous

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

- mathmath333

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

- Michele_Laino

hint:
this inequality:
\[\large \Large \frac{{10}}{{{x^2} - 5x + 16}} > 1\]
is equivalent to these systems:
\[\left\{ \begin{gathered}
10 > {x^2} - 5x + 16 \hfill \\
\hfill \\
{x^2} - 5x + 16 > 0 \hfill \\
\end{gathered} \right.\quad \cup \quad \left\{ {\begin{array}{*{20}{c}}
\begin{gathered}
10 < {x^2} - 5x + 16 \hfill \\
\hfill \\
\end{gathered} \\
{{x^2} - 5x + 16 < 0}
\end{array}} \right.\]

- anonymous

Simply taking one example from your answer.
If 2

- mathmath333

but wolfram gave this
http://www.wolframalpha.com/input/?i=+domain+f%28x%29%3D%5Clog_%7B10%7D+%5B1-%5Clog_%7B10%7D+%28x%5E2-5x%2B16%29%5D

- mathmath333

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

- anonymous

Oh wait, damn.

- anonymous

Ahahah.

- anonymous

Yeah, it's 2

- anonymous

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

- anonymous

Oh, got it.

- mathmath333

everything is correct except u graphed wrong

- anonymous

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

- anonymous

Big blunder on my side, I apologize.

- mathmath333

np thnks

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