Math help (2 Viewers)

[Fawun]

Question 1: g) and l) please

Thanks

... if you can see it lol because I don't know how to use latex

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

 

[Fawun]

Re: Math help (Because Carrot made me do this)

Thanks carrot but for part L, how did you get the answer? I don't get how you went from 4^n-2 x 3 blah blah to 1 over 4 x 3^4

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

 

[Fawun]

Re: Math help (Because Carrot made me do this)

idgi.

For the denominator for the first one, it's only 4^n-1 x 3^n+2 so how come in the second one, you have 4 x 4^n-2 x 3^n-2 x 3^4? like how come the 4^n-2 and 3^n-2 came out of nowhere?

Also for part g, the answer is

 

[enoilgam]

Re: Math help (Because Carrot made me do this)

idgi.

For the denominator for the first one, it's only 4^n-1 x 3^n+2 so how come in the second one, you have 4 x 4^n-2 x 3^n-2 x 3^4? like how come the 4^n-2 and 3^n-2 came out of nowhere?

Also for part g, the answer is

Carrot forgot to simplify. For l, 4^n-2 times 4 equals 4^n-1 (n - 2 + 1 = n - 1, the one from 4^1), its rearranged in order to allow for simplification. Similarily, carrot has broken 3^n+2 into 3^n-2 x 3^4 (n - 2 + 4 = n - 2).

 

[Fawun]

Re: Math help (Because Carrot made me do this)

New question that I need help with:

A lake has an initial pollution of 2 grams/cm^3. Everyday a river flows in a large amount of water of low pollution, and the same amount of water flows out to another river so that the amount of pollution (in gram/cm^3) in the lake at any day t is given by Q = 0.4 + Be^-0.02t

How long will it take to reduce the pollution level to 1 gram/cm^3?

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

Substitute t=0, Q=2, to find B.

2 = 0.4+Be^0

2 = 0.4+B

B=1.6

So therefore Q = 0.4 + 1.6e^-0.02t

We want the level to be 1, which means Q=1.

1 = 0.4 + 1.6e^-0.02t

0.6 = 1.6e^-0.02t

e^-0.02t = 3/8

-0.02t = ln(3/8)

t = -ln(3/8) / 0.02

t ~ 49.04

 

[Fawun]

Re: Math help (Because Carrot made me do this)

Substitute t=0, Q=2, to find B.

2 = 0.4+Be^0

2 = 0.4+B

B=1.6

So therefore Q = 0.4 + 1.6e^-0.02t

We want the level to be 1, which means Q=1.

1 = 0.4 + 1.6e^-0.02t

0.6 = 1.6e^-0.02t

e^-0.02t = 3/8

-0.02t = ln(3/8)

t = -ln(3/8) / 0.02

t ~ 49.04

Okay I got it thanks. Quick question, I don't get the relationship between log natural and normal logs lol. Like how do you change ln to log form? idgi.

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

You need to use the change of base law. Natural log is base e but 'normal log' is (I assume) base 10.

So ln = log_e whereas log (just by itself) is assumed to be log_10.

Not sure why you would want to change ln to log_10 since generally problems use ln as opposed to log_10.

But anyway to change it to base 10, we would do this:

 

[Fawun]

Re: Math help (Because Carrot made me do this)

You need to use the change of base law. Natural log is base e but 'normal log' is (I assume) base 10.

So ln = log_e whereas log (just by itself) is assumed to be log_10.

Not sure why you would want to change ln to log_10 since generally problems use ln as opposed to log_10.

But anyway to change it to base 10, we would do this:

Oh. So how would you go abouts in solving for x?

ln(x^2+1)=e ?

Do I expand it?

and what about evaluating without using a calculator:

e^ln 4+ ln 7?

do I have to change anything to log form lol

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

ln(x^2+1) = e = ln(1) (we do this so we can equate the inside)

x^2+1 = 1

x^2 = 0

x = 0.

Remember that e and ln are inverses of each other, so they 'cancel' each other out (function composition etc).

So e^ln(k) = k for any real k.

So therefore e^ln(4) = 4

So the final answer is 4 + ln(7).

 

[Fawun]

Re: Math help (Because Carrot made me do this)

ln(x^2+1) = e = ln(1) (we do this so we can equate the inside)

x^2+1 = 1

x^2 = 0

x = 0.

Remember that e and ln are inverses of each other, so they 'cancel' each other out (function composition etc).

So e^ln(k) = k for any real k.

So therefore e^ln(4) = 4

So the final answer is 4 + ln(7).

wat.

So whenever there is an ln and a e in a question, I can just cancel each other out?

So for this question,

ln2e^4 - lne^3, I can just cancel the ln and the e out? elaborate pls.

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

wat.

So whenever there is an ln and a e in a question, I can just cancel each other out?

So for this question,

ln2e^4 - lne^3, I can just cancel the ln and the e out? elaborate pls.

They ONLY cancel each other out if there's nothing else except e and ln.

For example:



And



But if I had:



It does NOT equal 2.

So for yor example ln2e^4 - lne^3:

ln(e^3) = 3 because there's nothing else.

But ln(2e^4) has to be changed to ln(2) + ln(e^4) = ln(2) + 4.

 

[Fawun]

Re: Math help (Because Carrot made me do this)

They ONLY cancel each other out if there's nothing else except e and ln.

For example:



And



But if I had:



It does NOT equal 2.

So for yor example ln2e^4 - lne^3:

ln(e^3) = 3 because there's nothing else.

But ln(2e^4) has to be changed to ln(2) + ln(e^4) = ln(2) + 4.

For the first example, how can you cancel the e and the ln if the ln and k are together?

For the second example, how can you cancel the ln and e when the e and k are together?

 

[Shadowdude]

Re: Math help (Because Carrot made me do this)

Because they're inverse functions.

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

For the first example, how can you cancel the e and the ln if the ln and k are together?

For the second example, how can you cancel the ln and e when the e and k are together?

Well you have to have something in the function f(x)=e^x.

You can't just hav a ln function by itself.

More specifically, (I know you don't like this math-y kind of explanation, but hopefully you will understand some part of it).

If f(x) and g(x) are inverse functions of each other, then f(g(x)) = g(f(x)) x.

So if f(x) = ln(x) and g(x) = e^x, then e^ln(x) = x and ln(e^x) = x.

 

[Fawun]

Re: Math help (Because Carrot made me do this)

Well you have to have something in the function f(x)=e^x.

You can't just hav a ln function by itself.

More specifically, (I know you don't like this math-y kind of explanation, but hopefully you will understand some part of it).

If f(x) and g(x) are inverse functions of each other, then f(g(x)) = g(f(x)) x.

So if f(x) = ln(x) and g(x) = e^x, then e^ln(x) = x and ln(e^x) = x.

Don't get any of what you just said. What is an inverse function? Maybe it is a good idea to **** learn my way through 3U next year since I obviously can't understand anything.

 

[Carrotsticks]

Re: Math help (Because Carrot made me do this)

Don't get any of what you just said. What is an inverse function? Maybe it is a good idea to **** learn my way through 3U next year since I obviously can't understand anything.

If you're going to get tutored, may as well go 1 to 1, not classes.

And no, please don't do that. It's not too late you know.

 

[Solution]

Re: Math help (Because Carrot made me do this)

For the first example, how can you cancel the e and the ln if the ln and k are together?

For the second example, how can you cancel the ln and e when the e and k are together?

For the second example, utilise your log laws:

ln(e^k) = kln(e)

Since we know ln(e) = 1 via our log laws,

ln(e^k) = k(1) = k.