Pre-Calculus Help: Solve x ≥ 0 and Log(base 3)x + Log(base 3)(x-6) = 3

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In summary: So at this point, your sign lines should look something like this:--0+++++++, with its zero directly beneath the "2"-mark on your number line, and the "+" sign to the right of it.5. You're almost done! Just beneath the "+" sign, write down the "sign line for (x-6) This will look like: -----0++++++, with its zero directly beneath the "6"-mark on your number line.6. Finally, beneath the "0" sign, write down the "sign line for (x+7): This will look like: --0+++++++, with its zero directly beneath the "7"-
  • #1
Help_Me_Please
19
0
This should be rather easy for the rest of you .. but somehow I can't remember what to do.

solve
[ (x-6) (x+7) ] / (x-2) is greater than or equal to 0

and

solve

log(base three)x + log(base three)(x-6) = 3

ty in advance =)
 
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  • #2
I tried solving the log equation and got to log(base three)x(x-6) = 3 and then forgot what to do @_@
 
  • #3
Help_Me_Please said:
I tried solving the log equation and got to log(base three)x(x-6) = 3 and then forgot what to do @_@

if
log(base a) p = q

then a^q = p

Can you take it from here?

Show what you tried for the first problem. I'd rather not just tell you how to do it. Do you have any ideas?
 
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  • #4
There are a few ways to solve the 1st problem. Here is the way I like doing it.

Get it in the form R(x) then inequality then 0, where R(x) is some rational function. You already have it in this form, but remember when you need to get it into this form you can't multiply or divide by a linear function of x.

Your critical values are:

x=-7
x=2
x=6

These are the values on the graph where the function changes sign. So look at less than x=-7, all the 3 factors are negative so the whole function is negative, each time you get to a critical value the sign changes. So for x<7 negative for -7<x<2 positive for 2<x<6 negative and for x>6 negative. Think what you want x to be and it's simple.
 
  • #5
I would like to add to Zurtex' good advice the following graphical procedure:
1. Draw the real numberline, and mark on it, at the very least, where each of the zeroes of your constituent polynomials are (in your case, your polynomials are all linear)
2. Directly beneath the drawn numberline, write down the "sign line for (x-6) like this: -----0++++++
The zero shall lie directly below the "6"-mark on your number line.
3. Continue to draw beneath your previous sign line the "sign line for (x+7):
This will look like: --0+++++++, with its zero directly beneath "-7".

4. You are to do the same with the (x-2) factor, the image you've drawn should look like this:
----------------------0+++++++++++++++
--0+++++++++++++++++++++++++++
----------------0+++++++++++++++++++
Now, you can at each point "multiply the signs", to form the "sign line for your rational function":
--0+++++++(inf)--0+++++++++++++++
(Note that as we approach the zero of the denominator, the function will tend to infinity (infinity sign depends on which direction you are approaching the singularity)
 
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FAQ: Pre-Calculus Help: Solve x ≥ 0 and Log(base 3)x + Log(base 3)(x-6) = 3

How do I solve for x in the given equation?

To solve for x, you first need to isolate the logarithmic terms on one side of the equation. In this case, you can combine the two logarithmic terms using the product rule of logarithms. Then, use the definition of logarithms to rewrite the equation in exponential form. Finally, solve for x by taking the appropriate root on both sides of the equation.

Why does x have to be greater than or equal to 0?

In this equation, the logarithm is only defined for positive values. If x is less than 0, the logarithm would be undefined. Therefore, x must be greater than or equal to 0 in order for the equation to be valid.

Can I use a calculator to solve this equation?

Yes, you can use a calculator to solve this equation. Most scientific calculators have a logarithm function (usually denoted as "log" or "ln") and an exponential function (usually denoted as "e^x" or "10^x"). You can use these functions to solve for x.

What is the significance of the number 3 in the logarithmic terms?

The number 3 is the base of the logarithm in this equation. This means that the logarithm is asking "what power do I need to raise 3 to, in order to get the value of x?" In other words, the equation is asking for the exponent in the exponential form of the equation.

What is the domain and range of this equation?

The domain of this equation is all real numbers greater than or equal to 0, since x must be positive. The range of the equation is all real numbers, since the logarithm can take on any value for a given input.

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