# Thread: Need help with a precalc question!

1. ## Need help with a precalc question!

hey guys, i have a question that i need help on.
it says use the "Rule of Four" to verbally, algebraicly, numerically, and graphically solve the problem.
Find all teh real numbers such that: x^2log(sub 3)x<5xlog(sub 3)x

anyone got any ideas to help me out with this?
i tried applying the change of base formula, but i dont know where i am going with that. please help

2. hey guys, i have a question that i need help on.
it says use the "Rule of Four" to verbally, algebraicly, numerically, and graphically solve the problem.
Find all teh real numbers such that: x^2log(sub 3)x<5xlog(sub 3)x

Ok, I have NO IDEA what the rule of four is... but I do have a solution.

$\displaystyle x^2log_3 x<5xlog_3x$

There is a $\displaystyle log_3 x$ on either side, so logically you would divide both sides by that, but when you do so, you have to consider the value of $\displaystyle log_3 x$ very carefully, if $\displaystyle log_3 x \ne0$, then you're fine, dividing both sides by $\displaystyle log_3 x$ gets you:

$\displaystyle x^2 < 5x$
$\displaystyle x^2 - 5x < 0$
$\displaystyle x(x-5) < 0$

When you plot the parabola $\displaystyle x(x-5)=0$, you can easily see that the solution for $\displaystyle x(x-5) < 0$ is $\displaystyle 0<x<5$

Now, consider the case when $\displaystyle log_3 x=0$, this only occurs if $\displaystyle x=1$:

When $\displaystyle x=1$, LHS = 0, RHS = 0, hence LHS is not less than RHS, which means if you solve $\displaystyle x^2log_3 x<5xlog_3x$, $\displaystyle x=1$ should NOT be a solution.

So the solution to $\displaystyle x^2log_3 x<5xlog_3x$ is $\displaystyle 0<x<5, x\ne1$.

Hope that helps.

3. Originally Posted by robdarftw
hey guys, i have a question that i need help on.
it says use the "Rule of Four" to verbally, algebraicly, numerically, and graphically solve the problem.
Find all teh real numbers such that: x^2log(sub 3)x<5xlog(sub 3)x

anyone got any ideas to help me out with this?
i tried applying the change of base formula, but i dont know where i am going with that. please help
1. Domain: $\displaystyle D=(0,\infty) \setminus \{1\}$

2. Since $\displaystyle \log_3(x) > 0 \ if\ x>1$

$\displaystyle x^2 \log_3(x)<5x \log_3(x)~\implies~x^2<5x~\implies~x(x-5)<0$. Therefore $\displaystyle 1<x<5$

3. Since $\displaystyle \log_3(x)<0\ if\ 0<x<1$

$\displaystyle x^2 \log_3(x)<5x \log_3(x)~\implies~x^2>5x~\implies~x(x-5)>0$. Therefore $\displaystyle x>5$

4. By the way: The case x = 5 is excluded because you'll get an equation if x = 5