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mathmath333
 one year ago
\(\large \color{black}{\begin{align}& \normalsize \text{Find the domain of the function} \hspace{.33em}\\~\\
& f(x)=\log_{10} [1\log_{10} (x^25x+16)] \hspace{.33em}\\~\\
\end{align}}\)
mathmath333
 one year ago
\(\large \color{black}{\begin{align}& \normalsize \text{Find the domain of the function} \hspace{.33em}\\~\\ & f(x)=\log_{10} [1\log_{10} (x^25x+16)] \hspace{.33em}\\~\\ \end{align}}\)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First thing's first  do you know what the domain of log (a) is ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Precisely! So from that we can deduce so much that 1  log (x^25x+16) has to be >0 for that whole thing to exist.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0From that we can deduce that log (x^25x+16) has to be <1 so that 1  log (x^25x+16) >0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But at the same time, for log (x^25x+16) to exist, (x^25x+16) also has to be >0.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2furthermore we have the subsequent condition for the existence of the inner logarithm: \[\Large {x^2}  5x + 16 > 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2then 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
 one year ago
Best ResponseYou've already chosen the best response.0It 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^25x+16) to exist, (x^25x+16) also has to be >0. We solve the quadratic we find that (x^25x+16) has complex roots and that means that (x^25x+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^25x+16) >0 We "flip" the log over a little and we have that 1 > log (x^25x+16) Express 1 as log (10) ( log(10) is 1 ) and you have log (10) > log (x^25x+16) Which equates to 10 > (x^25x+16) Or 0 > (x^25x+6) We solve this new quadratic and we have that (x^25x+6)=(x2)*(x3). What that means is that (x^25x+6) will look something like this: dw:1435346669499:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And to my guess, the quadratic itself will look something like this dw:1435347000458:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Therefore, for 0>x^25x+6, the quadratic must be negative and that only happens for x between (3,2).

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is the answer \(3<x<2\) or \(2<x<3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, it's the same thing really, solve either one of them and you'll get there.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large \left\{ \begin{gathered} \frac{{10}}{{{x^2}  5x + 16}} > 1 \hfill \\ \hfill \\ {x^2}  5x + 16 > 0 \hfill \\ \end{gathered} \right.\) by evaluating this i got \(2<x<3\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0Simply taking one example from your answer. If 2<x<3, take for x=2.5 x^25x+16=6.2512.5+16=9.75 10/9.75 is not greater than 1. Which means your evaluation is wrong.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but wolfram gave this http://www.wolframalpha.com/input/?i=+domain+f%28x%29%3D%5Clog_%7B10%7D+%5B1%5Clog_%7B10%7D+%28x%5E25x%2B16%29%5D

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1r u sure that \(10/9.75\) is not greater than \(1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, it's 2<x<3. flutter, where did I go wrong ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Big medal for whoever finds the point at which I went wrong with my asumptions.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1everything is correct except u graphed wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The roots of the quadratic were 2 and 3 not 2 and 3.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Big blunder on my side, I apologize.
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