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Algebra & Calculus questions

Hi, I start this thread to put up some various questions as time goes by to save starting a new thread with every question.

First one, what is the method I use to answer this

184∗625032000000=3a∗5b\displaystyle\dfrac{18^4 * 6250^3}{2000000} = 3^a * 5^b

Find values of a and b

hmmm, would I separate the fraction into two and cancel down?? No

Don't get what I am suposed to do, divide top and bottom by 2.

Unparseable latex formula:

\displaystyle\dfrac{9^4 * 3125^3}{1 * 10^6} = \dfrac{3^8 * 5^1^5}{1 * 10^6} = 3^a * 5^b

(edited 11 years ago)

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Reply 1
Avatar for Notnek
Notnek
21
A few hints:

Unparseable latex formula:

\displaysty18^4 = (2\cdot 3^2)^4=...



62503=(55×2)3=...\displaystyle 6250^3 = (5^5 \times 2)^3 = ...

2000000=2×106=2×(2⋅5)6=...\displaystyle 2000000 = 2 \times 10^6 = 2 \times (2\cdot 5)^6 = ...
Reply 2
Avatar for SubAtomic
SubAtomic
OP
11
Original post by notnek
A few hints:

184=(2⋅32)4=...\displaystyle 18^4 = (2\cdot 3^2)^4=...

62503=(55×2)3=...\displaystyle 6250^3 = (5^5 \times 2)^3 = ...

2000000=2×106=2×(2⋅5)6=...\displaystyle 2000000 = 2 \times 10^6 = 2 \times (2\cdot 5)^6 = ...


So I will write it like this and cancel down ??

Unparseable latex formula:

\displaystyle\dfrac{2^7\cdot 3^8\cdot 5^1^5}{2^7\cdot 5^6}



so a = 8 and b = 9
(edited 12 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by SubAtomic
So I will write it like this and cancel down ??

Unparseable latex formula:

\displaystyle\dfrac{2^7\cdot 3^8\cdot 5^1^5}{2^7\cdot 5^6}



so a = 8 and b = 9

That's correct.
Reply 4
Avatar for SubAtomic
SubAtomic
OP
11
Original post by notnek
That's correct.


Thanks, so is this just basic algebraic manipulation and being able to spot factors straight away? I saw the 2 * somethings but wasn't sure what to do. Cheers
Reply 5
Avatar for Notnek
Notnek
21
Original post by SubAtomic
Thanks, so is this just basic algebraic manipulation and being able to spot factors straight away? I saw the 2 * somethings but wasn't sure what to do. Cheers


I wouldn't say this question is basic. The first thing I did was look at the RHS and see that we require powers of 3 and powers of 5. Then I looked back to the LHS and noticed that the first term could be changed so that involves a power of 3 (since 18 is a multiple of 3), the second term could involve a power of 5 (6250 is a multiple of 5) and the denominator could also be changed to include a power of 5 (it's clearly a power of 10 which is a multiple of 5).

If you have this in your mind, then you have a direction to go in which makes the question easier.
(edited 12 years ago)
Reply 6
Avatar for SubAtomic
SubAtomic
OP
11
Original post by notnek
I wouldn't say this question is basic. The first thing I did was look at the RHS and see that we require powers of 3 and powers of 5. Then I looked back to the LHS and noticed that the first term could be changed so that involves a power of 3, the second term could involve a power of 5 and the denominator could also be changed to include a power of 5.

If you have this in your mind, then you have a direction to go in which makes the question easier.


Thank you, much appreciated.
Reply 7
Avatar for SubAtomic
SubAtomic
OP
11
Been a while but suppose that's good because haven't been stuck lol.

So I have an assignment and am having trouble with rearranging after have done the calculus or maybe it is before, anyway it is a differential equation so if someone can give me a few to do (logs, trig, etc) and maybe see if my methods are right because I cannot for the life of me manipulate my answer to look anything like the selection of answers ( I ask this because I cannot be accused of cheating this way ).

There is also a y = something when x = something part to it. So I think it is find the explicit/implicit type question.

This isn't it but am just showing what I do :s-smilie:

dydx=(x2y2)3/4=x6/4y6/4\displaystyle\dfrac{dy}{dx} = (x^2y^2)^{3/4} = x^{6/4}y^{6/4}

So

∫y−6/4dy=∫x6/4dx\displaystyle\int y^{-6/4} dy = \int x^{6/4} dx

Unparseable latex formula:

\displaystyle 2y^{-1/2} = - \left(\dfrac{2}{5}x^5^/^2 + C \right)



Take reciprocal ?

y2=−(52x−5/2+1C)\displaystyle\dfrac{ \sqrt y}{2} = - \left (\dfrac{5}{2}x^{-5/2} + \dfrac{1}{C} \right)


y=−2(52x−5/2+1C)\displaystyle \sqrt y = -2\left(\dfrac{5}{2}x^{-5/2} + \dfrac{1}{C} \right)

So

y=4(52x−5/2+1C)2\displaystyle y = 4\left(\dfrac{5}{2}x^{-5/2} + \dfrac{1}{C} \right)^2 ??


Think I may have created a bit of a tougher question than the one I am stuck on:s-smilie: So if anyone can give me an example like I asked above it'd be great.
(edited 11 years ago)
Original post by SubAtomic
Unparseable latex formula:

\displaystyle 2y^{-1/2} = - \left(\dfrac{2}{5}x^5^/^2 + C \right)



Take reciprocal ?

y2=−(52x−5/2+1C)\displaystyle\dfrac{ \sqrt y}{2} = - \left (\dfrac{5}{2}x^{-5/2} + \dfrac{1}{C} \right)

You were going fine until you did this. a=b+c  ⟹  1a≡1b+c≢1b+1ca=b+c \implies \dfrac{1}{a} \equiv \dfrac{1}{b+c} \not\equiv \dfrac{1}{b}+\dfrac{1}{c} (the latter being what you've wrongly implemented).
Reply 9
Avatar for SubAtomic
SubAtomic
OP
11
:s-smilie:

Unparseable latex formula:

\displaystyle\dfrac{ \sqrt y}{2} = - \dfrac{1}{\left(2 x^5^/^2\right)/5 + C}

??
Original post by SubAtomic
:s-smilie:

Unparseable latex formula:

\displaystyle\dfrac{ \sqrt y}{2} = - \dfrac{1}{\left(2 x^5^/^2\right)/5 + C}

??

Yep, and this can be tidied up slightly by writing C=25KC=\dfrac{2}{5}K (which is fine as C is arbitrary), multiplying through the whole thing by 2 and multiplying numerator and denominator of the resulting RHS by 52\dfrac{5}{2}.
Reply 11
Avatar for SubAtomic
SubAtomic
OP
11
Original post by Farhan.Hanif93
Yep, and this can be tidied up slightly by writing C=25KC=\dfrac{2}{5}K (which is fine as C is arbitrary), multiplying through the whole thing by 2 and multiplying numerator and denominator of the resulting RHS by 52\dfrac{5}{2}.


Will do this a step at a time because am not sure

y=(−225(x5/2+K))\displaystyle \sqrt y = \left(-\dfrac{2}{\frac{2}{5}(x^{5/2} + K)}\right) ??

Couldn't work out how to tex this and have the 2/5 factor on the bottom:colondollar:
(edited 11 years ago)
Original post by SubAtomic
Will do this a step at a time because am not sure

Unparseable latex formula:

\displaystyle \sqrt y = \left(-\dfrac{2}{2x^5^/^2/5 + 2K/5}\right)

??

Couldn't work out how to tex this and have the 2/5 factor on the bottom:colondollar:

Can you cancel the factors any further? (think about the 2's). What about multiplying top and bottom by 5?
Reply 13
Avatar for SubAtomic
SubAtomic
OP
11
Ok so

Unparseable latex formula:

\displaystyle 5 \cdot \sqrt y = \left(-\dfrac{10}{2\left(x^5^/^2 + K \right)}\right)}



Unparseable latex formula:

\displaystyle = -\dfrac{5}{\left(x^5^/^2 + K\right)}



Or am I way off ?
(edited 11 years ago)
Original post by SubAtomic
Ok so

Unparseable latex formula:

\displaystyle 5 \cdot \sqrt y = \left(-\dfrac{10}{2\left(x^5^/^2 + K \right)}\right)} = -\dfrac{5}{x^5^/^2 + K}



Or am I way off ?

That's right but you can still simplify it further. You have 5⋅y=102⋅(−1x52+K)5\cdot y = \dfrac{10}{2} \cdot \left(- \dfrac{1}{x^{\frac{5}{2}} + K}\right).

Simplify 10/2 and the 5's cancel.
Reply 15
Avatar for SubAtomic
SubAtomic
OP
11
Original post by Farhan.Hanif93
That's right but you can still simplify it further. You have 5⋅y=102⋅(−1x52+K)5\cdot y = \dfrac{10}{2} \cdot \left(- \dfrac{1}{x^{\frac{5}{2}} + K}\right).

Simplify 10/2 and the 5's cancel.


Shocking, I always forget this on the spot, one day:rolleyes:

Unparseable latex formula:

\displaystyle \left(-\dfrac{10}{2\left(x^5^/^2 + K \right)}\right)}

=102⋅(−1x52+K)\displaystyle = \dfrac{10}{2} \cdot \left(- \dfrac{1}{x^{\frac{5}{2}} + K}\right)

Forgetting to take the factor out makes things look very different:cool:

Thanks for your help will rep when I can again.

Care to give me a trig or log or something example to work on and come back later with?
(edited 11 years ago)
Original post by SubAtomic
Shocking, I always forget this on the spot, one day:rolleyes:

Unparseable latex formula:

\displaystyle \left(-\dfrac{10}{2\left(x^5^/^2 + K \right)}\right)}

=102⋅(−1x52+K)\displaystyle = \dfrac{10}{2} \cdot \left(- \dfrac{1}{x^{\frac{5}{2}} + K}\right)

Forgetting to take the factor out makes things look very different:cool:

Thanks for your help will rep when I can again.

Care to give me a trig or log or something example to work on and come back later with?

I haven't put much thought in but you should be able to try:

dydx−y2cos⁡x=0\dfrac{dy}{dx}-y^2 \cos x =0.

dydx=cos⁡2ysin⁡x\dfrac{dy}{dx} = \cos^2y \sin x

dydx=xy(y−1)\dfrac{dy}{dx} = xy(y-1)

Harder one: dydx=(1−y2)x1−x2\dfrac{dy}{dx} = \dfrac{(1-y^2)x}{\sqrt{1-x^2}}
(edited 11 years ago)
Reply 17
Avatar for SubAtomic
SubAtomic
OP
11
1. dydx−y2cos⁡x=0\dfrac{dy}{dx}-y^2 \cos x =0.

So

dydx=y2cos⁡x\dfrac{dy}{dx} = y^2 \cos x

1y2 dy=cos⁡x dx\dfrac{1}{y^2} \ dy = \cos x \ dx

Integrate

∫y−2 dy=∫cos⁡x dx\int y^{-2} \ dy = \int \cos x \ dx

−1y=sin⁡x+C - \dfrac{1}{y} = \sin x + C

Reciprocal then times -1

y=−1sin⁡x+C\displaystyle y = -\frac{1}{\sin x + C}


2. dydx=cos⁡2ysin⁡x\dfrac{dy}{dx} = \cos^2y \sin x

So

∫1cos⁡2ydy=∫sin⁡xdx\displaystyle\int\dfrac{1}{\cos^2y} dy = \int \sin x dx

tan⁡y=−cos⁡x+C \tan y = - \cos x + C ?

Can I divide by tan to get y alone or is this where I stop?

3. dydx=xy(y−1)\dfrac{dy}{dx} = xy(y-1)

#EDIT Should have seen this, Cheers Raheem.

So divide through y(y – 1) first?

1y(y−1) dy=x dx\displaystyle\dfrac{1}{y(y-1)} \ dy = x \ dx

Integrate

∫1y2−y dy=∫x dx\displaystyle\int\dfrac{1}{y^2-y} \ dy = \int x \ dx

ln⁡∣y−1∣−ln⁡(y)=x22+C\displaystyle \ln |y-1| - \ln (y) = \dfrac{x^2}{2} + C

Can I take this further?

4. dydx=(1−y2)x1−x2\dfrac{dy}{dx} = \dfrac{(1-y^2)x}{\sqrt{1-x^2}}

Hmmm had a go but think I may be going wrong somewhere.

So I multiplied by the denominator, did reciprocal, multiplied by x and that is where stopped as am not sure if am going wrong

(1–x2)1/2dydx=(1–y2)x\displaystyle (1 – x^2)^{1/2} \dfrac{dy}{dx} = (1 – y^2)x

x1–x2 dx=11–y2 dy\displaystyle \dfrac{x}{\sqrt{1–x^2}} \ dx = \dfrac{1}{1–y^2} \ dy

Right or wrong??

Best stop there because I may be wrong on more than one of these questions:rolleyes:
(edited 11 years ago)
Reply 18
Avatar for raheem94
raheem94
13
Original post by SubAtomic

2. dydx=cos⁡2ysin⁡x\dfrac{dy}{dx} = \cos^2y \sin x

So

∫1cos⁡2ydy=∫sin⁡xdx\displaystyle\int\dfrac{1}{\cos^2y} dy = \int \sin x dx

tan⁡y=−cos⁡x+C \tan y = - \cos x + C ?

Can I divide by tan to get y alone or is this where I stop?


Your solution to question 1 is correct.

For question 2, take the tan inverse of both sides,
tany=−cosx+C  ⟹  arctan(tany)=arctan(−cosx+C)  ⟹  y=arctan(−cosx+C) \displaystyle tany=-cosx+C \implies arctan(tany)=arctan(-cosx+C) \\ \implies y=arctan(-cosx+C)
(edited 11 years ago)
Reply 19
Avatar for raheem94
raheem94
13
Original post by SubAtomic


3. dydx=xy(y−1)\dfrac{dy}{dx} = xy(y-1)

So divide through (y – 1) and y first?

(1y−1/y)dy=xdx\displaystyle\left(\dfrac{1}{y-1}/y\right) dy = x dx

Rearrange and integrate

∫(1y−1+1)dy=∫xdx\displaystyle\int \left(\dfrac{1}{y-1} + 1\right) dy = \int x dx

y+ln⁡∣(y−1)∣=x22+C\displaystyle y + \ln |(y-1)| = \dfrac{x^2}{2} + C

Can I take this further?



You have gone completely wrong here.
dydx=xy(y−1)  ⟹  ∫1y(y−1) dy=x dx \displaystyle \frac{dy}{dx} = xy(y-1) \implies \int \frac{1}{y(y-1)} \ dy=x \ dx

Use partial fractions to simplify the LHS,
1y(y−1)≡Ay+By−1 \displaystyle \frac{1}{y(y-1)} \equiv \frac{A}{y} + \frac{B}{y-1}

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