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Old 04-27-2016, 05:18 AM   #1
sickufully
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Default Indices, Exponential and Logarithmic Equations + Trig + Derivatives

Yeah so this trimester I'm taking Math Methods (not by choice obviously) and since I pretty much failed what we call 'Further Maths' (the level below Methods) in yr12, I really don't see myself passing this subject unless I receive some assistance. There are some questions which have really confused me and aren't covered in the text book so if anyone knows how to solve these, please share your wisdom.

The following questions are the ones I'm having trouble with. I've attempted them but my answers just don't look right. I may post more if I get stuck again.

PS - Sorry if I didn't set my working out correctly, I'm fairly new to posting math homework online.

Indices


So for this, I wasn't sure if I first needed to switch the x^-1 and the y^-1 in the bracket and change them to positive indices OR if I took the 3 from outside the brackets and added it to each indice within the bracket. So I didn't very far with this one.


Exponential


For this, I basically just applied simple algebra rules and multiplied the first bracket by the second. So I got:
e^2x * e^2x
e^2x * -e^-x
e^-x * e^2x
e^-x * -e^-x

Collecting terms, I ended up with: e^4x -e^x

Not even sure if I did it correctly, probably wasn't even the right rule to use.

Logarithm


This isn't overly hard, I just got confused because I'm used to seeing one of the logs as just lnX. E.g. What I'm used to seeing: 3ln2 + 2ln3 - ln4.

So I've got:
= ln(2^3) + ln(3^2) - ln(4^3)
= ln8 + ln9 - ln64
= ln8 * ln9 = ln72
= ln72/ln64 = ln9/8

Am I anywhere near the answer?



I've done a question like this before but I didn't fully understand it. This is what I have so far:
= ln[(2x)^3)] - ln(4x^4) + ln[(3x^2)^2)]
= ln(8x^3) - ln(4x^4) + ln(9x^4)

I don't know what to do next, divide ln(8x^3) - ln(4x^4) OR multiply ln(4x^4) + ln(9x^4).

I think I'm supposed to divide first so I end up with ln(2x^-1) + ln(9x^4) but that doesn't seem right.

My answer ended up being: ln18x^3



This last one really threw me off as I'm not used to having the 2 in front of the log. Looking back through all lecture notes and examples, there's nothing that covers numbers in front of log.

Am I supposed to multiply something by 2 or divide by 2 or multiple the whole equation or...? WHAT DOES THIS MEAN!?
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Old 04-27-2016, 07:13 AM   #2
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Default Re: Indices, Exponential and Logarithmic Equations

i'm not sure how math methods works (it could be about calculation efficiency or something) but any solution should be fine, so long as all the steps are right. though i guess a general rule of thumb is to always try to work with small numbers and as few things as possible, so there's less chance of messing something up (and not noticing it later)

first one: i'd say flip the indices first, because then the x and y terms can combine and there'll be less terms to exponentiate
2x^2 * (x^-1*y^2/2xy^-1)^3
= 2x^2 * (y^2*y^1/2x*x^1)^3
= 2x^2 * (y^3/2x^2)^3
= 2x^2 * y^9 / 8x^6
= y^9 / 4x^4

second one: the algebra's a little off but expanding it out is what i'd do
(e^2x + e^-x)*(e^2x - e^-x)
= e^2x*e^2x - e^2x*e^-x + e^-x*e^2x - e^-x*e^-x
= e^4x - e^x + e^x - e^-2x
= e^4x - e^-2x

alternatively the product's of the form (a+b)(a-b) = a^2-b^2 [where a=e^2x, b=e^-x] and you can multiply it like this:
(e^2x + e^-x) * (e^2x - e^-x) = (e^2x)^2 - (e^-x)^2 = e^4x - e^-2x
but both methods are equally good

third one: you got it

fourth one: yeah i'd divide cause it'd keep the x indice small
3 ln(2x) - ln(4x^4) + 2 ln(3x^2)
= ln[(2x)^3)] - ln(4x^4) + ln[(3x^2)^2)]
= ln(8x^3) - ln(4x^4) + ln(9x^4)
= ln(2x^-1) + ln(9x^4)
= ln(18x^3)

there's a little trip-up with multiplying too because of the negative sign on ln(4x^4): if you multiply you'd actually be calculating ln(8x^3) - [ln(4x^4) + ln(9x^4)] = ln(8x^3) - ln(4x^4) - ln(9x^4). i guess you could imagine brackets around the group you want to divide/multiply, and see if that's equivalent to the equation you're trying to simplify (or avoiding negatives in general)

if you wanna double-check your answer you can use wolframalpha or another algebra tool (some of them have step-by-step generators too, though the method they use might not be the prettiest)

anyways, hope that helps

edit: congrats on 666 posts :P

Last edited by beary605; 04-27-2016 at 07:14 AM..
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Old 04-27-2016, 08:28 AM   #3
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Default Re: Indices, Exponential and Logarithmic Equations

First one:
Order of operations, parenthesis first (transpose the x and y), then apply the cubic outside.


Second one:
This is a difference of two squares, so you are left with
e^4x - e^2x.


Third one:
You're right.

Fourth one:
beary knows what's up.
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Old 04-27-2016, 05:47 PM   #4
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Default Re: Indices, Exponential and Logarithmic Equations

Man, I can't express how much I love you guys. I don't know how this stuff actually stays in your head, I forget it as soon as the topic is over.

I probably should've mentioned that Methods is just a level of math we have. It goes something like Foundation Math > Further Maths > Math Methods > General Maths > Algorithmics. I struggled with Further in yr12 and now 5 years later, I have to do Methods... GAH!

Anyway,
First one - once you told me what step to do first, I was able to work the rest out fairly easily. So flipping the indices in the bracket (if negative) should always be the first step, mostly for ease of working out the rest of the equation?

Second one - With the law you guys posted, I can't seem to find it in the text book nor in any lecture notes. Knowing my study habits, I probably just skipped over it but do you mind if I ask where you got that from?

Third one - I'm a friggen genius.

Fourth one - So I was kinda heading in the right direction but for some reason I thought you couldn't have negative indices so the dividing of the first two ln's seemed incorrect to me. I also keep forgetting that for multiplying indices, you just add them but the base is still multiplied. It just gets confusing :/

Fifth one - Still not sure what to do. Without the 2 in front of the log, its just log4^2=16. I'm just unsure what step comes next when dealing with the 2.

Thanks again for your time and expertise beary and rushy, really appreciate it. I may post more in the next day or two as I have an assignment coming up next week for this.

edit - rip 666 posts :P
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Old 04-27-2016, 07:16 PM   #5
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Default Re: Indices, Exponential and Logarithmic Equations

fifth: what are you getting hung up on? the 2? it looks like you've already been using alog(x) = log(x^a), so basically proceed as normal

base 4 2log(16)
= base 4 log(16^2)
= base 4 log(256)
= 4
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Old 04-27-2016, 08:40 PM   #6
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Default Re: Indices, Exponential and Logarithmic Equations

Yeah the 2 in front of it just throws me off. I wasn't sure if you had to multiply the whole equation by 2 or just the 16. Or something else. I knew it was probably really simple but I couldn't find any examples. So I really appreciate your answer, cheers

But basically, how I interpret the way you did it:
- Square the 16 (16^2) to get rid of the 2
- New equation is log4(256)
- So now I just need to work out what 4^x = 256
- Which = 4
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Old 04-27-2016, 08:52 PM   #7
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Default Re: Indices, Exponential and Logarithmic Equations

Quote:
Originally Posted by sickufully View Post
Second one - With the law you guys posted, I can't seem to find it in the text book nor in any lecture notes. Knowing my study habits, I probably just skipped over it but do you mind if I ask where you got that from?
Difference of Squares is an elementary factoring method learned in Algebra 1.
https://www.khanacademy.org/math/alg...nce-of-squares
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Old 04-27-2016, 08:57 PM   #8
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Default Re: Indices, Exponential and Logarithmic Equations

yeah, flipping negative indices in a fraction is good, they'll usually cancel some term out or combine with something, and make the expression smaller/easier to work with

numbers in front always imply multiplication, like 4x^2, 3 sin x, so for the fifth one you calculate
2*log4 16 = 2*2 = 4
bringing the number in works too but the numbers get messy for something like 10 log3 27
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Old 04-27-2016, 09:38 PM   #9
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Default Re: Indices, Exponential and Logarithmic Equations

Quote:
Originally Posted by beary605 View Post
yeah, flipping negative indices in a fraction is good, they'll usually cancel some term out or combine with something, and make the expression smaller/easier to work with

numbers in front always imply multiplication, like 4x^2, 3 sin x, so for the fifth one you calculate
2*log4 16 = 2*2 = 4
bringing the number in works too but the numbers get messy for something like 10 log3 27
Yeah most of the questions we're tasked with require us to provide our answer with no negative indices. So thanks for that advice.

And for your answer to the fifth question, you basically just worked it out as if the 2 wasn't there, and then just multiplied your answer by 2?

So:
- log4(16) = 2 * 2 = 4

Or another example:
- 3log7(49)
= log7(49) = 2 * 3 = 6
OR
- 3log7(49)
= log7(49^3)
= log7(117649) = 6


Quote:
Originally Posted by rushyrulz View Post
Difference of Squares is an elementary factoring method learned in Algebra 1.
https://www.khanacademy.org/math/alg...nce-of-squares
Yeah I realised how simple it is. I was kinda heading in the right direction, but I should probably brush up on simple rules like that one. Cheers for the link too
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Old 04-28-2016, 06:27 AM   #10
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

I'm not fully understanding what it's asking me to do here. I do understand the 4th quadrant must have cos𝜃>0 but I'm unsure if I'm applying the following rule correctly (I have no idea for the second one (ii), when tan𝜃<0):

Rule - cos^2𝜃 + sin^2𝜃 = 1



My attempt went something like this for (i):
Cos^2𝜃 + Sin^2𝜃 = 1
Cos^2𝜃 + (-3/5)^2 = 1
Cos^2𝜃 + (9/25) = 1
Cos^2𝜃 + (9/25) - (9/25) = 1 - (9/25)
Cos^2𝜃 = (25/25) - (9/25)
Cos^2𝜃 = (16/25)
Cos𝜃 = sqrt(16/25)
Cos𝜃 = ħsqrt(16/25)
Cos𝜃 = ħ(4/5)
Cos𝜃 = 4/5 (because if it's in the 4th quadrant, it has to be positive)

Am I applying the rule correctly or am I way off?
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Old 04-28-2016, 07:00 AM   #11
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

create a right triangle by drawing a terminal arm in the fourth quadrant and connecting a line from the end to the x-axis.

sin theta = y/r

therefore y = -3 and r = 5

you can use pythagorean theorem to get x or just realize it's a 3,4,5 triangle thus x=4

cos theta = x/r = 4/5

then, tan theta = y/x = -3/4

basically you did it right but there are much easier ways

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Old 04-28-2016, 08:11 AM   #12
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

You make it look so easy! We just have to show working out, hence why I wrote quite a few steps. Cheers for the clarification though. If possible, using the lengthy method I posted, would you or anyone able to write the first line for the (ii) question for me?

If I was to say: Tan^2𝜃 + Sin^2𝜃, would I be correct?
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Old 04-28-2016, 09:36 AM   #13
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

A geometric proof should be sufficient, as explained by dynam0. As long as you have a right triangle with an angle and a side, you can derive all 3 sides and all 3 angles of the triangle, meaning you can solve for cos(theta) and tan(theta).

The method you described is correct for an algebraic proof, however this problem lends itself to geometry very well.
As per your request of the first line of (ii), there is only one line required. You have sin(theta) and cos(theta) already, and you know that tan = sin/cos, therefore:

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Old 04-28-2016, 09:54 AM   #14
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

Something like this:


You will need to negate certain terms due to it being in the fourth quadrant, but I can't do everything for you
EDIT: I've practically done it already, so w/e: the sine and tangent terms are negative.
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Old 04-29-2016, 07:20 AM   #15
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

Yeah pretty sure I was over-thinking this. I was trying to use tan^2𝜃 = sin^2𝜃/cos^2𝜃 and this just confused me with the ^2 and I was getting weird fractions. Basically I was trying to use the method I used for cos𝜃 but this is how I ended up solving it:

tan𝜃 = sin𝜃/cos𝜃
tan𝜃 = -3/5 / 4/5
tan𝜃 = -3/5 * 5/4
tan𝜃 = -15/20
tan𝜃 = -3/4

Thanks for your explanations + diagrams rushy and dynamo, appreciate it.
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Old 04-29-2016, 04:46 PM   #16
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Default Re: Indices, Exponential and Logarithmic Equations + Trig Question

np broheim
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Old 04-29-2016, 10:22 PM   #17
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Default Re: Indices, Exponential and Logarithms Equations + Trig + Derivatives

I may or may not have one more question. But it's on derivatives. I'll assume you guys also know this stuff since you seem to know everything else.

We have to find dy/dx for each of the following.



Rules
if y = sinkx, then dy/dx = kcoskx
if y = coskx, then dy/dx = -ksinkx

The first one (vi) - I think it goes something like this:

y = 6sin(x/3) + 3cos(2x/3)
dy/dx = 6*(1/3)cos(x/3) + 3*(-2x/3)sin(2x/3)
dy/dx = 3/1cos(x/3) - 6x/3sin(2x/3)

Can I then simplify further to get:
dy/dx = 3cos(x/3) - 3x/1sin(2x/3)
or do I need to keep k as a fraction?



Rules
if y = e^kx, then dy/dx = ke^kx
if y = x^n, then dy/dx = nx^n-1

The second one (vii) - I didn't get very far. I think the first step is as follows, but I'm unsure of what to do next.

y = 3/x + e^3x/3 - 1/x^2
dy/dx = 3/1 + 3e^3x/3 - 1/2x
dy/dx = ???
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Old 04-29-2016, 11:02 PM   #18
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Default Re: Indices, Exponential and Logarithmic Equations + Trig + Derivatives

Quote:
Originally Posted by sickufully View Post
The first one (vi) - I think it goes something like this:

y = 6sin(x/3) + 3cos(2x/3)
dy/dx = 6*(1/3)cos(x/3) + 3*(-2x/3)sin(2x/3)
dy/dx = 3/1cos(x/3) - 6x/3sin(2x/3)

Can I then simplify further to get:
dy/dx = 3cos(x/3) - 3x/1sin(2x/3)
or do I need to keep k as a fraction?
You're on the right track here. The first half is totally correct, however it looks like you forgot to differentiate 2x/3 for the cosine part, that's the only problem. Oh, and 6/3 = 2, not 3


Quote:
Originally Posted by sickufully View Post
Rules
if y = e^kx, then dy/dx = ke^kx
if y = x^n, then dy/dx = nx^n-1

The second one (vii) - I didn't get very far. I think the first step is as follows, but I'm unsure of what to do next.

y = 3/x + e^3x/3 - 1/x^2
dy/dx = 3/1 + 3e^3x/3 - 1/2x
dy/dx = ???
3/x does not differentiate to 3/1, it would if x were in the numerator. If you leave x in the denominator, you will need to use the quotient rule which is quite messy, which is why we move them to the numerator by using negative exponents. If you do not know this rule, I suggest you brush up on it! You've differentiated the middle term perfectly.

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Old 04-29-2016, 11:55 PM   #19
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Default Re: Indices, Exponential and Logarithmic Equations + Trig + Derivatives

So with the first one, when you do this step:


What exactly happens to the 6 out the front and the +3 before (-sin...)?
Are they just divided by each other, so the 6 = 2 and the 3 = 1 (which just disappears)?

And for the second one, I completely missed that rule for 1/x = x^-1. Also, the final simplifying step you did, is it necessary to do that or is it OK to leave it as: -3/x^2 + e^3x + 2/x^3?

I do understand how you got 2-3x/x^3 + e^3x, I'm just wondering if I'll lose marks in a test for not simplifying further.
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Old 04-30-2016, 01:05 AM   #20
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Default Re: Indices, Exponential and Logarithmic Equations + Trig + Derivatives

Quote:
Originally Posted by sickufully View Post
So with the first one, when you do this step:


What exactly happens to the 6 out the front and the +3 before (-sin...)?
Are they just divided by each other, so the 6 = 2 and the 3 = 1 (which just disappears)?
6 * 1/3 = 2, 3 * 2/3 = 2. Distributive property moves the 2 outside the parenthesis.

Quote:
Originally Posted by sickufully View Post
And for the second one, I completely missed that rule for 1/x = x^-1. Also, the final simplifying step you did, is it necessary to do that or is it OK to leave it as: -3/x^2 + e^3x + 2/x^3?

I do understand how you got 2-3x/x^3 + e^3x, I'm just wondering if I'll lose marks in a test for not simplifying further.
You won't lose points on a test if you don't simplify it all the way down like that, but you'll look like one smart cookie if you do~
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