Logarithms

I would use the change of base theorem to write:

[MATH]\log_9(x)=\frac{\log_3(x)}{\log_3(9)}=\frac{\log_3(x)}{2}[/MATH]
Can you proceed?
 
Problem 2 is trigonometry; in particular, you can use the formula for tan(a - b).

Problem 3 is also trigonometry; there are several ways you could do this one, including calculus and another angle-difference identity, namely for sin(a - b).

None of this is pre-algebra!

Do you have any work to show? Or is the name of the topic all you need?
 
Sooooo...

Which thread do you want your responses in? This one or the other in Pre-Algebra?

-Dan
 
The two threads were about different questions on the same paper, one in algebra and two in trig.
 
In the original thread he was asking about questions 1, 2, and 3. In the second thread he simply said that question 1 is challenging, leading me to believe that he wanted help on just question number 1.

-Dan
 
topsquark said:
In the original thread he was asking about questions 1, 2, and 3. In the second thread he simply said that question 1 is challenging, leading me to believe that he wanted help on just question number 1.

-Dan
Click to expand...
I actually need help on all three, I've never come across the questions, I'm a pure science student so I wasn't thought. I'm also new here so I didn't really know you could see all the threads. And also I posted them with the hope of getting the topics precisely, so I can read them up. That is asides getting the problems solved.
 
If you apply the theorem I suggested, you then have:

[MATH]\log_3(6)-\frac{\log_3(x)}{2}=\frac{1}{2}[/MATH]
Next, I would observe that:

[MATH]\log_3(6)=\log_3(2\cdot3)=\log_3(2)+\log_3(3)=\log_3(2)+1[/MATH]
And so we now have:

[MATH]\log_3(2)+1-\frac{\log_3(x)}{2}=\frac{1}{2}[/MATH]
Collect like terms:

[MATH]\log_3(2)-\frac{\log_3(x)}{2}=-\frac{1}{2}[/MATH]
Multiply by 2:

[MATH]2\log_3(2)-\log_3(x)=-1[/MATH]
Apply some rules for logs

[MATH]\log_3(4)-\log_3(x)=-1[/MATH]
[MATH]\log_3\left(\frac{4}{x}\right)=-1[/MATH]
[MATH]\log_3\left(\frac{x}{4}\right)=1[/MATH]
Convert from logarithmic to exponential form:

[MATH]\frac{x}{4}=3[/MATH]
[MATH]x=12[/MATH]
 
Any progress on the trig problems, #2 and #3? Do you know the sum and difference identities? Have you tried applying them?

(This is one reason for submitting problems separately rather than putting them all in one thread, where some can get lost.)
 
MarkFL said:
If you apply the theorem I suggested, you then have:

[MATH]\log_3(6)-\frac{\log_3(x)}{2}=\frac{1}{2}[/MATH]
Next, I would observe that:

[MATH]\log_3(6)=\log_3(2\cdot3)=\log_3(2)+\log_3(3)=\log_3(2)+1[/MATH]
And so we now have:

[MATH]\log_3(2)+1-\frac{\log_3(x)}{2}=\frac{1}{2}[/MATH]
Collect like terms:

[MATH]\log_3(2)-\frac{\log_3(x)}{2}=-\frac{1}{2}[/MATH]
Multiply by 2:

[MATH]2\log_3(2)-\log_3(x)=-1[/MATH]
Apply some rules for logs

[MATH]\log_3(4)-\log_3(x)=-1[/MATH]
[MATH]\log_3\left(\frac{4}{x}\right)=-1[/MATH]
[MATH]\log_3\left(\frac{x}{4}\right)=1[/MATH]
Convert from logarithmic to exponential form:

[MATH]\frac{x}{4}=3[/MATH]
[MATH]x=12[/MATH]
Click to expand...
Thank you,
So the topic is just Logarithms or is it advanced ?
 
Dr.Peterson said:
Any progress on the trig problems, #2 and #3? Do you know the sum and difference identities? Have you tried applying them?

(This is one reason for submitting problems separately rather than putting them all in one thread, where some can get lost.)
Click to expand...
Yes on Number 2, i tried using an engineering textbook to read trigonometry, but number 3, I couldn't even start.
 
Treymor said:
Yes on Number 2, i tried using an engineering textbook to read trigonometry, but number 3, I couldn't even start.
Click to expand...

So, I take it these problems are not from a course you are taking, but from something you haven't learned at all (prep for some test?). What is the context of your questions? (We ask you to state this in our submission guidelines.)

You really need to take a course, or carefully study a book, rather than ask individual questions as if you can jump into the middle of a course. These are relatively challenging trig problems.

But if you think you should be able to handle these, please show your work on #2, and whatever ideas you do have for #3, and we can discuss them. We need to know what you do know, in order to help effectively.

As for #1, I would not call that very advanced. It would probably be covered in an algebra course that introduces logs.
 
Yes, Its a question from an exam past question I came across, I want to study it and that's why I requested for the topics.
 
This is the solution to number 2
 

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Good -- so you do know enough trig!

In fact, your solution is much nicer than what I did (just to check that the problem made sense) -- I replaced beta with (pi/4 - alpha) in the expression and just simplified, where you evidently either saw that the angle-sum identity could be manipulated to obtain that expression, or just played with it until you discovered something that worked. Any of these is a reasonable thing to do; typically the "elegant" solution (like yours) is not the first solution one comes up with, but is found after the fact by looking back at the answer and the work.

How about trying #3 now. My answer is not pretty (yet); maybe you actually have done something reasonable, but just don't think it looks good enough. You might just try substituting as I described for #2, in order to obtain a nicer form from which the max and min can be found; or you might see something more direct. I will tell you that I came at it knowing that a sum of phase-shifted sines will result in a different shifted sine, so that the answer will just be the amplitude of that sinusoid; that gave me a general direction to go. If I expected you to be able to handle basic calculus, I would have tried that, which may be easier.

Now, you haven't really answered my context question. Is this a past exam in a course you are taking, or for a contest or admission exam, or what? That may make a difference in what kind of studying to suggest, as well as in what to expect of you.
 
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