IB test questions

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Leah5467

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Feb 28, 2019
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Hello, i don't know if my answers are correct in the test and i am very worried. These are the questions:
1.f(x)=2sinx,given maximum point=2 and minimum point=-2,
find the greatest value of "p" in 2sin(px) if it cannot cross y<-1 at any point

2.Solve x for log4(2x)=log16(13x-4)

Thank you! It would ease my worries if i at least know where i am at.
 
I can't help you with Question 1, because it doesn't make sense to me. "p" affects the period not the amplitude so the max of 2 sin(px) will still be 2 and min will be -2 for any value of p, so the graph will go below y=-1. Are you sure you have the question correct? Was it perhaps 2sin(x) + p ?

With Question 2, I would rewrite both sides using the "change of base" rule and go from there.
 
Leah5467 said:
... find the greatest value of "p" in 2sin(px) if [its graph] cannot cross y<-1 ...
Click to expand...
Hi. I agree with Harry_the_cat; the exercise statement seems flawed.

y < -1 describes a region in the xy-plane, not a line. So the graph of 2sin(px) won't "cross" it.

This question makes sense: "For what value(s) of p does the graph of y=2sin(px) not enter the region y<-1?"

I doubt that's the question they tried to ask, but its answer is zero.

?
 
Leah5467 said:
... 2. Solve [for x]

log4(2x)=log16(13x-4)
Click to expand...
Here are two requests from the forum's submission guidelines:

(1) Show what you've already tried or thought about

(2) Start new threads for new exercises


I agree with Harry_the_cat's suggestion: Use the Change-of-Base formula.

I would use it to convert the right-hand side from log16 to log4.

That way, you can use a property of logs to put the given equation into this form:

log4(expression1) = log4(expression2)

For that equation to be true, expression1 must equal expression2.

But, what approach did you try?

?
 
Harry_the_cat said:
I can't help you with Question 1, because it doesn't make sense to me. "p" affects the period not the amplitude so the max of 2 sin(px) will still be 2 and min will be -2 for any value of p, so the graph will go below y=-1. Are you sure you have the question correct? Was it perhaps 2sin(x) + p ?

With Question 2, I would rewrite both sides using the "change of base" rule and go from there.
Click to expand...
Thank you!
Welp it has the graph on the paper,but i don't have the paper with me. The roots are pie and 2pie,like any normal sine graph. The maximum and minimum points are -2 and 2.
F(x) intersects the line y=-1 at 7pie/6 and 11pie/6. But any point in the sine model cannot intersect with y=-1.

I do it using this method:
I figure if the lowest point of the graph does not touch y=-1,then the whole graph won't touch the line.
So i set up this:

2sin(px)<-1
2sin(p(270 degree))<-1
Sin(p(270 degree))<-1/2
px<220 degree
And i get p<7/9

The log one i used change of base:
Log4(2x)=log4(13+4x)/log416

Log44x2=log4(13+4x)

4x2=13+4x
And i solve this by quadratic formula because i cannot calculate roots from cross method. But golly,i am so scared because the calculations are quite a lot for a non-calculator test. Maybe i did something wrong?
 
Leah5467 said:
... [the test question] has the graph on the paper, but i don't have the paper with me ...

... i get p<7/9
Click to expand...
We'll have to wait until you get the actual question in hand. (Your descriptions and work have issues.)

By the way, it's easy to check some values of p smaller than 7/9 and see they don't work. For example, google the following and look at the graph.

plot 2*sin(1/9*x) , -1


... i used change of base:
Log4(2x) = log4(13+4x)/log416

Log4(4x2) = ...
Click to expand...
That is good. Next, what is log4(16) ?

That is not good. We can't replace 2x with 4x^2 because those two expressions are not the same for all x.

Once you replace log4(16) with its value, then consider the following property of logarithms.

n log(a) = log(an)

?
 
Thank you! What i was trying to do was:2log42x,can it be log4(2x)^2? Gosh,i am scared...
 
Leah5467 said:
2,and then can i times the 2 to ther other side?
Click to expand...
Oh, I see what went wrong before. My check of your result in post #5 failed because I didn't notice your typos (you typed 13x+4 throughout, instead of 13x-4). I didn't realize you had multiplied both sides by 2 and applied the property; I would have, had you answered my question in post #8. (This is just one reason why the forum's guidelines ask students to not skip our questions.)

Sorry I contributed to your nervousness. You were on the right track! With corrections (and showing steps), you have:

Log4(2x) = log4(13 - 4x)/2

2Log4(2x) = log4(13 - 4x)

Log4(4x2) = log4(13 - 4x)

4x2 = 13x - 4

Solving that gives the solutions. Good job.

?
 
Otis said:
Oh, I see what went wrong before. My check of your result in post #5 failed because I didn't notice your typos (you typed 13x+4 throughout, instead of 13x-4). I didn't realize you had multiplied both sides by 2 and applied the property; I would have, had you answered my question in post #8. (This is just one reason why the forum's guidelines ask students to not skip our questions.)

Sorry I contributed to your nervousness. You were on the right track! With corrections (and showing steps), you have:
Log4(2x) = log4(13 - 4x)/2

2Log4(2x) = log4(13 - 4x)

Log4(4x2) = log4(13 - 4x)

4x2 = 13x - 4

Solving that gives the solutions. Good job.

?
Click to expand...
Yep I got \(\displaystyle 4x^2 - 13x + 4 = 0\) also, and then used the quadratic formula to solve for x.
 
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