Question

\(\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}}\)

Answer 1

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

Answer 2

a>0

Answer 3

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.

Answer 4

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

Answer 5

ok

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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: