## mathmath333 one year ago \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}}

1. anonymous

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

2. mathmath333

a>0

3. 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.

4. 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

5. mathmath333

ok

6. anonymous

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

7. Michele_Laino

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

8. Michele_Laino

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

9. Michele_Laino

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

10. 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|

11. anonymous

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

12. anonymous

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

13. mathmath333

is the answer $$-3<x<-2$$ or $$2<x<3$$

14. anonymous

. . .

15. anonymous

-3<x<-2

16. anonymous

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

17. 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 $$2<x<3$$

18. Michele_Laino

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

19. anonymous

Simply taking one example from your answer. If 2<x<3, take for x=2.5 x^2-5x+16=6.25-12.5+16=9.75 10/9.75 is not greater than 1. Which means your evaluation is wrong.

20. mathmath333
21. mathmath333

r u sure that $$10/9.75$$ is not greater than $$1$$

22. anonymous

Oh wait, damn.

23. anonymous

Ahahah.

24. anonymous

Yeah, it's 2<x<3. flutter, where did I go wrong ?

25. anonymous

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

26. anonymous

Oh, got it.

27. mathmath333

everything is correct except u graphed wrong

28. anonymous

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

29. anonymous

Big blunder on my side, I apologize.

30. mathmath333

np thnks

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