1. ## Questions

Wow, I can't believe I've forgotten how to do these, but I have. Any help would be appreciated.

1) Simplify 7root(x^9)/5root(x^6)

2) Solve the equation both algebraically and graphically (I've done it the graphical way, cannot seem to figure out how to do it algebraically)

3) Rewrite the expression logbase5(x+3) into an equivalent expression using only natural logarithms.

4) Describe the transformations that can be used to transform the graph of log(x) to a graph of f(x) = 4log(x+2)-3.

5) Solve the equation 2sin^2(x)cos(x)=cos(x) algebraically

6) Find all the exact solutions to 2sin^2(x)+3sin(x)-2=0 on the interval [0,2pi)

7) Use a graphing calculator to solve the following for x. e^(2x)=3x^2

I graphed it in my graphing calculator, but I am not sure if you're trying to find the intersection of the two graphs or what? I found the intersection and plugged it into x, but they are not equal, so I am not sure what to do.

8) State the smallest interval (
0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)

By the way, I am not asking you to solve these outright, but just help me. These problems are from PreCalculus, but I just forget how to solve them. Thank you for your time!

2. ## A different approach

Hello zaga04

Welcome to Math Help Forum!
Originally Posted by zaga04
Wow, I can't believe I've forgotten how to do these, but I have. Any help would be appreciated.

1) Simplify 7root(x^9)/5root(x^6)

2) Solve the equation both algebraically and graphically (I've done it the graphical way, cannot seem to figure out how to do it algebraically)

3) Rewrite the expression logbase5(x+3) into an equivalent expression using only natural logarithms.

4) Describe the transformations that can be used to transform the graph of log(x) to a graph of f(x) = 4log(x+2)-3.

5) Solve the equation 2sin^2(x)cos(x)=cos(x) algebraically

6) Find all the exact solutions to 2sin^2(x)+3sin(x)-2=0 on the interval [0,2pi)

7) Use a graphing calculator to solve the following for x. e^(2x)=3x^2

I graphed it in my graphing calculator, but I am not sure if you're trying to find the intersection of the two graphs or what? I found the intersection and plugged it into x, but they are not equal, so I am not sure what to do.

8) State the smallest interval (
0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)

By the way, I am not asking you to solve these outright, but just help me. These problems are from PreCalculus, but I just forget how to solve them. Thank you for your time!
May I suggest that you'll be more likely to get help if you (a) post one question at a time (see Rule 14); (b) show us that you've made some sort of attempt, so that we can help you at the specific point that you're having a problem with.

Having said that, here are some suggestions for the first couple of questions.

(1) I take it that you mean: Simplify $\displaystyle \frac{\sqrt[7]{x^9}}{\sqrt[5]{x^6}}$.

You need to use two rules of indices:

• $\displaystyle \sqrt[a]{x^b} = x^{\frac{b}{a}}$

• $\displaystyle \frac{x^a}{x^b}=x^{a-b}$

So $\displaystyle \sqrt[7]{x^9}=x^{\frac{9}{7}}$, and $\displaystyle \sqrt[5]{x^6}=...$ ?

Then you'll need to subtract the second power of $\displaystyle x$ from the first.

(2) $\displaystyle |4x-3| = 5 \sqrt{x+4}$

$\displaystyle \Rightarrow (4x-3)^2 = 25(x+4)$

$\displaystyle \Rightarrow 16x^2 - 24x +9 = 25x +100$

$\displaystyle \Rightarrow ...$ ?

Solve the quadratic equation. (I don't think it factorises, so you'll have to use the formula.)

3. Sorry about posting so many questions, thought it would be easier than posting so many threads. Thanks for helping me out for the first two problems, make sense. I'll come back with the work I've done for the others.

4. So I think I got #3, log5(x+3) converted to ln would = ln(x+3)/ln5

Would #4 be like the graph of 4log(x+2)-3 is vertically stretched by 4, moved horizontally 2 units to the left, and moved vertically down 3 units? I am not sure if this is what they are asking.

I am not sure how to start #5, #6 or #8 at all. #8 I graph it on my calculator, but I'm not sure what the smallest interval would be that would give the complete graph of the polar equation.

#7 I already explained what I did, I graphed it on my graphing calculator, but I am not sure how you find x from the graphing calculator? Do you find the intersection of the two graphs e^(2x) and 3x^2?

That's the thought process I've put into these problems.

5. Originally Posted by zaga04
So I think I got #3, log5(x+3) converted to ln would = ln(x+3)/ln5

Would #4 be like the graph of 4log(x+2)-3 is vertically stretched by 4, moved horizontally 2 units to the left, and moved vertically down 3 units? I am not sure if this is what they are asking.

I am not sure how to start #5, #6 or #8 at all. #8 I graph it on my calculator, but I'm not sure what the smallest interval would be that would give the complete graph of the polar equation.

#7 I already explained what I did, I graphed it on my graphing calculator, but I am not sure how you find x from the graphing calculator? Do you find the intersection of the two graphs e^(2x) and 3x^2?

That's the thought process I've put into these problems.
Q3: Yes, those two expressions are equal (change of base rule)

Q4: It would be assuming the original graph is log(x)

Q5: Take cos(x) from both sides

$\displaystyle 2sin^2(x)cos(x) - cos(x) = 2sin^2(x)(cos(x)-1) = 0$

and if ab=0 either a=0 or b=0. Normally this kind of question will have limits on solutions because there are an infinite number.

Q6: Let u = sin(x) and solve the quadratic

Q7: Yep that's what you do

$\displaystyle e^{2x} = 3x^2$ cannot be solved algebraically hence the calculator

Q8: It's asking how much space you need to view a complete wave of the equation. Remember cos(ax) is the graph of cos(x) squashed by a factor of a and that cos(x) is periodic over 2pi

6. Thank you so much
For #5, I understand the process, but what kind of answer would you get?

For #6, I got pi/6 and 5pi/6.

For #7, I got the intersection to be x=-.39.

For #8, I got the interval to be (0≤theta≤pi)

Are the ones I did correct?

7. Originally Posted by zaga04
Thank you so much
For #5, I understand the process, but what kind of answer would you get?

For #6, I got pi/6 and 5pi/6.

For #7, I got the intersection to be x=-.39.

For #8, I got the interval to be (0≤theta≤pi)

Are the ones I did correct?
5. $\displaystyle sin^2(x) = 0$ or $\displaystyle cos(x) = 1$

$\displaystyle x = k \pi \: ,\, k\, \in \, \mathbb{Z}$

In this case sin^2(x) is 0 when cos(x) is also 0.

6,7. Put the values you found back into the equation and see if it works out to be true

8. ## Polar equation sketch

Hello zaga04
Originally Posted by zaga04
...
8) State the smallest interval (
0 ≤theta ≤k) that gives the complete graph of the polar equation r=4cos(5theta)
To sketch the polar graph, note that the important values of cosine occur at intervals of $\displaystyle \pi/2$:

$\displaystyle \cos 0 = 1,\,\cos(\pi/2) = 0,\, \cos(\pi) = -1, \,\cos(3\pi/2)=0,\,\cos(2\pi) = 1, ...$

These will occur when $\displaystyle \theta = 0, \pi/10,\, 2\pi/10,\, 3\pi/10,...$

So sketch a diagram with radial lines every $\displaystyle \pi/10$; i.e. $\displaystyle 18^o$ from $\displaystyle 0^o$ to $\displaystyle 180^o$.

Mark the point $\displaystyle (4,0)$. The curve starts here and comes in to the origin with the line $\displaystyle \theta = \pi/10$ as a tangent. Then it continues through negative values of $\displaystyle r$ (i.e. out the other side into the third quadrant) going out to $\displaystyle (-4, \pi/5)$; comes back to the origin tangential to $\displaystyle \theta = 3\pi/10$; etc.

See the attached sketch.

You'll find that you need $\displaystyle 0 \le \theta \le \pi$ to get the complete curve.

Also I was trying to rework #5, would this be right?

$\displaystyle 2sin^2(x)cos(x)=cos(x)$
$\displaystyle 2sin^2(x)=1$
$\displaystyle sin^2(x)=\frac{1}{2}$
$\displaystyle sin(x)=\sqrt\frac{1}{2}$
$\displaystyle x=\arcsin\sqrt\frac{1}{2}$
$\displaystyle x=\frac{\pi}{4}$

10. ## The two great commandments

Hello zaga04
Originally Posted by zaga04

Also I was trying to rework #5, would this be right?

$\displaystyle 2sin^2(x)cos(x)=cos(x)$
$\displaystyle 2sin^2(x)=1$
$\displaystyle sin^2(x)=\frac{1}{2}$
$\displaystyle sin(x)=\sqrt\frac{1}{2}$
$\displaystyle x=\arcsin\sqrt\frac{1}{2}$
$\displaystyle x=\frac{\pi}{4}$
There are two great commandments of mathematics. The first is this:

• Thou shalt not divide by zero.

and the second is:

• When thou takest a square root, forget not thy plus or minus sign.

And I'm afraid you've broken them both!

You've got to be careful you don't divide both sides of an equation by something that might have the value zero. If you do, you might miss possible solutions.

In $\displaystyle 2\sin^2x\cos x = \cos x$, where there's a factor of $\displaystyle \cos x$ on both sides, note that $\displaystyle \cos x = 0$ is a possible solution - it makes both sides have an equal value, $\displaystyle 0$. So before you divide by $\displaystyle \cos x$, say that $\displaystyle \cos x =0$ or ... then do the division. So you get:

$\displaystyle 2\sin^2x\cos x = \cos x$

$\displaystyle \Rightarrow \cos x = 0$ or $\displaystyle 2\sin^2x = 1$

The first possibility gives us $\displaystyle x = \frac{\pi}{2}, \frac{3\pi}{2}, ...$

And the second: $\displaystyle \sin^2x = \frac12$

So don't forget the second commandment:

$\displaystyle \sin x = \pm\frac{1}{\sqrt2}$

So you not only get $\displaystyle x = \frac{\pi}{4},\frac{3\pi}{4}, \frac{9\pi}{4}, ...$ corresponding to $\displaystyle \sin x = +\frac{1}{\sqrt2}$, but also the odd multiples of $\displaystyle \frac{\pi}{4}$ in between:

$\displaystyle x=\frac{5\pi}{4}, \frac{7\pi}{4}, ...$

which come from $\displaystyle \sin x = -\frac{1}{\sqrt2}$

So don't forget to obey the commandments!

11. Thank you again Grandad and as you can tell I am not very good at math. I was just trying to do it by myself, but thank you for correcting me and telling me the right way.

So the dots at the end of each answer would correspond that you have to keep on adding multiples of $\displaystyle \frac{\pi}{4}$? Or as the answer on my sheet can I just leave it as ... at the end?

12. Hello zaga04
Originally Posted by zaga04
Thank you again Grandad and as you can tell I am not very good at math. I was just trying to do it by myself, but thank you for correcting me and telling me the right way.

So the dots at the end of each answer would correspond that you have to keep on adding multiples of $\displaystyle \frac{\pi}{4}$? Or as the answer on my sheet can I just leave it as ... at the end?
Yes, the three dots symbol, ..., is called an ellipsis (not to be confused with an ellipse, which is a squashed circle), and it indicates that things have been left out.

You can use it to indicate a finite sequence, like $\displaystyle 2, 4, 6, ..., 10$, or an infinite one, like $\displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, ...$ which continues without ending.