Algebra question

[clear roost]

What i need to do and how to sloved this?

[clear roost]

<@&286206848099549185>

[slender veldt]

Substitute $u := e^{2x}$

[blazing wolf]

adonhs

[slender veldt]

And since $u\neq 0$ you can multiply the equation by u on both sides, then you get a quadratic equation in terms of u.

[blazing wolf]

adonhs

[clear roost]

Whats U?

[slender veldt]

a variable

[clear roost]

And then

What we need to do

[slender veldt]

Like I didn't tell

[clear roost]

I mean what the answer want?

[slender veldt]

You have to calculate the solutions of the equation:

$\ 3e^{2x} - 4e^{-2x} = 5\$

Substitute with $u = e^{2x}$:

$\ 3u - 4u^{-1} = 5\$

$\ \Rightarrow 3u - \frac{4}{u} = 5\$

Multiply by u both sides since $u = e^{2x} \neq 0$

$\ \Rightarrow 3u^2 - 4 = 5u\$

$\ \Rightarrow 3u^2 -5u - 4 = 0\$

Resubstitute back after you got the solutions.

[blazing wolf]

adonhs

[clear roost]

Ok i has been get it

What the answer

I try to check my answer

[slender veldt]

What's your answer? Show it.

[clear roost]

Wait hold on

I got wrong

Is that possible to factoring that?

[clear roost]

Oh ya I get a question, why 4u^-1 its 4/u not 1/4u^1

[clear roost]

i got the answer is

5 plus minus sqrt -23/6

[slender veldt]

Kinda now resubstitute and solve for x:

$\e^{2x} = u_1 = \frac{5+\sqrt{73}}{6}\$

Notice that $u_2 = \frac{5-\sqrt{73}}{6} < 0\$

So that is not a solution.

[blazing wolf]

adonhs

[clear roost]

wait

my answer is sqrt -23

[slender veldt]

This cant be

[clear roost]

hmm ok

so is cant be use a quadratic formula??

[slender veldt]

You can

You just probably did a miscalculation

[clear roost]

oh

[slender veldt]

That can happen

[clear roost]

ok let me check

[slender veldt]

$3u^2 -5u -4 = 0\$

$\ \Rightarrow u_{1,2} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4\cdot 3 \cdot (-4)}}{2\cdot 3}\$

[blazing wolf]

adonhs

[clear roost]

yeah thats right

same as me

[slender veldt]

As you can see there is no negative value under the square root

minus times minus is positive

[clear roost]

ohh

i see

i wrong about this

it will be 25 + 48 right?

[slender veldt]

no worries happens

Yes

[clear roost]

damn im not in focus

[slender veldt]

then take a rest lol

[clear roost]

after this haha

ok i has been get it 5 plus minus sqrt 73/6

[slender veldt]

yea and we only have to consider the plus case

because the minus case gives us a negative value

and we know that the exponential function is always positive in this case

[slender veldt]

reference

[clear roost]

why it can be like that

cuz some question have like x1=3 ,x2=-5

but in this case must be +

[slender veldt]

because exponential function has positive values only

$\ e^{2x} > 0 \quad \forall x \in \mathbb{R}$

[blazing wolf]

adonhs

[slender veldt]

So there is no x for a negative value

[clear roost]

what is R

e^2x > 0 forr all x is an element of R

[slender veldt]

Yes

R is real numbers

[clear roost]

ohh ok

hmm

bro what must im doing

Do I have to break this root form

llike simplification

[slender veldt]

you have to solve for x now

Resubstitute and solve for x:

$\e^{2x} = u_1 = \frac{5+\sqrt{73}}{6}\$

Notice that $u_2 = \frac{5-\sqrt{73}}{6} < 0\$

So that is not a solution since $e^{2x} > 0 \quad \forall x \in \mathbb{R}$

[blazing wolf]

adonhs

[clear roost]

Ok if (u) is : e^2x

It will be

e^2x = 5+√73/6
&
e^2x = 5-√73/6

But how to solve X

Its will be like

e^x = 5±73/6

Bro im stuck at there

[slender veldt]

ln() om both sides

$ln(e^{2x}) = 2x \cdot ln(e) = 2x$

[blazing wolf]

adonhs

[slender veldt]

And then you almost already solved it

[clear roost]

What is ln

Oh integral?

[slender veldt]

inverse function of e^x

[slender veldt]

no what

you dont know ln(x) and e^x

?

logarithms

[clear roost]

Ohhh i know that

[slender veldt]

$ln(x) = log_{e}x$

[blazing wolf]

adonhs

[clear roost]

Yess

[slender veldt]

You should know that they cancel out

$ln(e^x) = log_{e}e^x = e^{ln(x)} = x$

[blazing wolf]

adonhs

[slender veldt]

I hope you check it

ankara messi

[clear roost]

🤣

I didnt get it bro

Btw i get this question from A level test Cambridge

😂

[slender veldt]

dont you understand that they cancel out?

[clear roost]

I didnt know about this logarithm
I just know about characteristic of logarithm

[slender veldt]

Suppose f(x) and its inverse function $f^{-1}(x)$.

Then it follows:

$\f^{-1}(f(x)) = x = f(f^{-1}(x))$

[blazing wolf]

adonhs

[slender veldt]

This is a charaterisric for all functions with their inverse

Example:

$\sin(arcsin(x)) = x\$

$\cos(arccos(x)) = x\$

$\ln(e^x) = x\$

[blazing wolf]

adonhs

[clear roost]

Damn i didnt learn that, i will learn more of this

Thanks bro i apreciated that, i will solved this by learn more one days

[slender veldt]

Yeah watch a video about this as it makes it solving equations easier

[clear roost]

Yes sir thanks bro

[slender veldt]

what's that?

[regal hearth]

just ln the entire equation

put ln next to all those numbers