logarithm

[vagrant badger]

what have i done wrong?

[astral coyote]

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[still acorn]

This should be a plus.

[vagrant badger]

oh ok

but why?

[still acorn]

Because log(ab) = log(a) + log(b), not log(a)log(b).

[vagrant badger]

you are right

so (x/5)^(ln x -2) is somehow the “b”?

[still acorn]

Yes.

[vagrant badger]

okok thanks

[vagrant badger]

what can i do now?

[still acorn]

Expand the brackets, some stuff should be able to cancel.

[vagrant badger]

but do i have to multiply the two brackets together?

how do i expand them?

[still acorn]

Well, as usual.

(a + b)(c + d) = ac + ad + bc + bd

[vagrant badger]

oh okok

can i just write this like this or can i do more:
lnx *(-ln 5)

[vagrant badger]

what do u think?

[still acorn]

For now, leave it as -ln(x) ln(5). You'll use it to factor the whole thing later.

So, what did you get?

[vagrant badger]

but shouldn’t the ln5 be negative?

[still acorn]

Well, it doesn't matter where you put the minus sign. Multiplication is commutative: a(-b) = (-a)b = -ab.

[vagrant badger]

oh true

well if thats so

this is what i have till now:

wait lnx* -ln5 should all be in brackets right?

[still acorn]

Alright.
Now, you can see that you have some similar terms.

[vagrant badger]

yess

[still acorn]

Well, it would look nicer if it did, but not necessary.

[vagrant badger]

okok

[vagrant badger]

i can do something with these

right

but can i do anything more

[still acorn]

Well, do that first.

[vagrant badger]

i feel something is odd

or..?

[still acorn]

I think you canceled something you shouldn't have. Let's see.

[vagrant badger]

ahah

definitely

[still acorn]

We have:
3ln(5) + ln(x)^2 - ln(5)ln(x) + 2ln(x) - 2ln(5) = 3ln(x)
First, let's group the ln(5) terms and ln(x) terms (apart from the one with the coefficient -ln(5)).
ln(x)^2 - ln(5)ln(x) + (3ln(5) - 2ln(5)) + (2ln(x) - 3ln(x)) = 0
ln(x)^2 - ln(5)ln(x) + ln(5) - ln(x) = 0

This is what we should get.

[vagrant badger]

is what u got the same as this:

[tough minnow]

I don't know if this is the method you choose to do but there are much faster approaches

[still acorn]

Well, now try to factor the resulting expression.

[still acorn]

Oh, like what?

[tough minnow]

well (x/5)^a = (x/5)^b

a=b

we just need to get the bases equal and equate the powers

[still acorn]

Ohh, right! I didn't even notice that.

That does make it a lot faster, yeah.

[tough minnow]

👍

[still acorn]

Well, we can't quite do that immediately when the base is a function, but taking the logarithm of both sides does allow us to factor it a lot faster.

[tough minnow]

oh

[vagrant badger]

is this correct?

[still acorn]

Yes.
Now, try grouping pairs of terms and factoring.

[vagrant badger]

can i write lnx as x and ln5 as y to make it easier to group pairs of terms and factoring?

[still acorn]

Well, you have to pick one. You can't call different things the same name.

I recommend taking y = ln(x).

[vagrant badger]

ok so lnx is “y” and ln5 can be “a” for example

[still acorn]

Sure, why not.

[still acorn]

No, don't factor y from three separate places.

[vagrant badger]

okok

so what should i do here

[still acorn]

Notice the following:
y^2 + a - ay - y = 0
(y^2 - ay) - (y - a) = 0

So, what can you do next?

[vagrant badger]

how did u separate it like that?

[still acorn]

I just grouped terms.

[vagrant badger]

i am not sure how to do this

[still acorn]

Again, try grouping terms as I did above.

[vagrant badger]

like so?

but (y-1) and (1-y) is that the same?

[still acorn]

No, but 1 - y = -(y - 1).

So, using that, you can now factor.

[vagrant badger]

aa ok

[still acorn]

Yeah, nice!

So, just substitute y and a back and you're done.

[vagrant badger]

finallyyy ahaha

[vagrant badger]

yess

but i have one last question

[vagrant badger]

what happened to the minus?

i forgot

to include it

or can i just leave it

[vagrant badger]

what do u think?

[still acorn]

That's not a minus.

That's multiplication.

[vagrant badger]

minus multiplied with the numbers in the bracket isnt it?

[still acorn]

Oh, that's what you meant.

You don't write it like that. Write it as -a(y - 1).

[vagrant badger]

aaa

understood

[vagrant badger]

so thats also why we got (y-a)

[still acorn]

Yes.

[vagrant badger]

ok thank uuu

[still acorn]

You're welcome!

I do recommend Shwifty's approach, by the way. It's a lot faster.

[vagrant badger]

i will leave the channel open for a while so i can take some notes

[vagrant badger]

ah ok i will take a look on it later

[vagrant badger]

i am doing task 8b now

and i have to use the result from a to solve the equation

which equals to 0

but can i do substitution here

lnx = u

ln5 = ?

[still acorn]

No need.
If uv = 0, then u = 0 or v = 0. So, you have two possibilities:
ln(x) - ln(5) = 0
ln(x) - 1 = 0
So, solve each one.

[vagrant badger]

aah i always forget that i can do this

thaanks

[vagrant badger]

like so?

[vagrant badger]

+close