Question

Can't follow maths in my lectures whatsoever

Answer 1

Alright, in the comments section, left-hand side, you should see an equation button. That's where you write it out.

Answer 2

\[Y = e ^{\tan2x}\]

Answer 3

Alright, cool. So are you trying to take the derivative of this, or what exactly do you want to do with it?

Answer 4

what's a derivative?

Answer 5

we're doing like trig differentiation and derivatives but like I've no clue what's going on and idk what "e" or "ln" are...

Answer 6

Alright, e is just a constant. It's a number, not a function, long story short, it doesn't matter for our purposes how or what it came from; it's a constant, and it's about 2.71

ln is very closely related to e. Natural log (ln) is something you can do to a number or variable, that puts that number in terms of e to some power, for instance,

Answer 7

\[\ln(e) = 1.\]What this means is that the argument of that natural log function, the e inside the parantheses, you are asking what power e is to. Since e inside the parentheses is to the first power, ln(e) = 1. On the other hand,
\[\ln(e^2) = 2\]\[\ln(e^3)=3\]\[\ln(e^4)=4\]Make sense?

Answer 8

No? what's a parentheses?

Answer 9

Parentheses are these: ( )

Answer 10

So it's just whatever inside the parentheses is what we'd call the "argument" of the natural log function.

So the argument of \[\ln(25)\]is 25. The argument of \[\ln(x^2)\] is x^2.

Answer 11

Ok

Answer 12

Alright, so your function is \[y = e^{\tan(x)}\]
In the same way that the argument of \[\ln(x)\] is x, the argument of \[\tan(x) \] in your problem is x; the thing inside the parentheses is your argument.

So, let me ask something: What's the derivative of \[y = x^2\] with respect to x?

Answer 13

I don't know

Answer 14

Alright, no worries. Do you know about the "limit definition of a derivative", or what a derivative comes from? This is the best place to start so everything makes sense in the longterm.

Answer 15

No

Answer 16

I did none of this at school

Answer 17

Alright, site is back.

Answer 18

so \[\ln (e ^{2x}) = 2x\]

Answer 19

Exactly, yup.

Answer 20

Alright, so the derivative is a special kind of limit. Do you know what limits are?

Answer 21

\[4\ln(5x+4) = 12.8 \] Use the law of logs to calculate for the value of x

Answer 22

Alright, the first thing you can do is ______ across. What can you do with the four and the 12.8 on the other side to simplify things, first off?

Answer 23

?

Answer 24

divide both by 4?

Answer 25

?

Answer 26

Yup. Alright, then you have \[\ln(5x+4) = 3.2\]

The question asks you to use the laws of logarithms to solve it, so from what you know about this problem, you can change things around like this:
\[e^{3.2} = 5x+4\]This is called exponentiation. To exponentiate something is to put e to the power of (whatever you're exponentiating). e.g.

If I said, "exponentiate the following equation",\[\ln(20) = x^2\]exponentiating it would result in \[e^{\ln(20)} = e^{x^{2}}\]

In the same odd way that when you take the natural log of e, you get one, because what you're really asking when you say, "take the natural log of (argument)" is, "what power of e is the argument to?", when you put \[e^{\ln(x)}\]the e^ln disappears, and you're left with the argument of the natural log function.

Answer 27

Does this make sense? \[e^{\ln(y)} = y\]\[e^{\ln(200)} = 200\]\[e^{\ln(2014)} = 2014\]

Answer 28

Jesus he made is so much more complex than it actually is..

Answer 29

Lmao.

Answer 30

Alright, so recap, in this second problem, you started with\[4\ln(5x + 4) = 12.8\]The first thing we did was choose to divide through by four on both sides to get:\[\ln(5x+4) = 3.2\]Now, we've just exponentiated the whole equation, so we have:\[e^{\ln(5x+4)} = e^{3.2}, \ \ 5x+4 = e^{3.2}\]Now, you can solve for x as you might usually. Give it a shot.

Answer 31

5x+4 = ln(e^3.2) ?

Answer 32

Wait a moment, you'd have to do that to both sides. Kind of like multiplying or dividing, if you exponentiate one side of an equation, you need to exponentiate both. Right now, that won't be very helpful, so just using plain old algebra should help. Get 5x by itself, and then get x by itself.

Answer 33

well how was I meant to know that?

Answer 34

Idk wtf you do with e

Answer 35

No worries, lol. It's like any other math operation; multiplication, division, addition, it always affects both sides, we just don't often realize it.

No problem, e is just a constant, not a variable, so you can treat it exactly like you would any other number.

So e might as well be 12, or 3. You can move it around; That being said, what do you think you should do to get x by itself?

Answer 36

I don't know?

Answer 37

idk what constants or variables are

Answer 38

Alright, so the next step here would be this:
\[5x = e^{3.2} - 4\]This make sense?

Alright, can talk about that in a second.

Answer 39

I'm actually so fluttering stupid

Answer 40

ha you can't swear. you can tell this site isnt irish

Answer 41

Lul, no problem. So after subtracting across, what do you think the next step is to get x completely by itself?

Answer 42

divide by 5

Answer 43

?

Answer 44

Yup! There you go. And now, all you have is constants on the right-hand side of your equation, so you can solve for x as a number. Were you allowed to use a calculator on this exam?

Answer 45

no

Answer 46

Was it multiple choice, or write-in?

Answer 47

writing. I think you can leave the equation as it is because a lot of what we were doing was just isolating x

Answer 48

Yep, exactly, so your final answer should be: \[x = \frac {e^{3.2}-4}{5}\]

Answer 49

ok same as my own

Answer 50

Alright, great!

Answer 51

what is the medal for? lol

Answer 52

Just because, haha, you figure it out.