Integrated Circus - 2021-04-08

@NicolásCastellanos Because! You always need to find both solutions if the equation is a quadratic. Sometimes, there will be no real solutions. Sometimes there will be two identical solutions, eg $(x-3)^2=0$. In some problems, one of the solutions may be invalid, eg because you only want positive solutions, or integer solutions. But you need to find those solutions before you decide what to do with them.

Nicolás Castellanos

8:46 PM

with minus, x=log_3(4)

or 1.26

robjohn

No, they said you need to find both solutions of the quadratic. but then they said some of the solutions might be invalid.

PM 2Ring

And remember, a similar thing happened when we had $z=\pm\sqrt{125}$. We could ignore the negative solution because $z=x^2$

Nicolás Castellanos

next equation

$3^{3x+1}=9\left(2^{x+3}\right)$

the right side can be simplified

to 18^(x+3)

then we can express in base 3

because 3*6 = 18

nono

18 cant be expressed with base 3

robjohn

@NicolásCastellanos no

Nicolás Castellanos

$3^{3x+1}=18^{x+3}$

robjohn

8:51 PM

$9\left(2^{x+3}\right)=3^2\cdot2^{x+3}$

Nicolás Castellanos

why?

robjohn

because $9=3^2$

Nicolás Castellanos

ok

wait

robjohn

you cannot simply say that $a\cdot b^c=(ab)^c$

Nicolás Castellanos

now the right side

make the multiplication

6^(2x+6)

robjohn

8:54 PM

what are you doing?

Nicolás Castellanos

multiplying the right side

robjohn

by what?

Nicolás Castellanos

making that multiplication

3^2 * 2^(x+3

robjohn

no you are not

The right side is $3^2\cdot2^{x+3}$

Nicolás Castellanos

so what must i do?

robjohn

8:56 PM

You cannot combine them any further. That is not $a^xa^y=a^{x+y}$

Nicolás Castellanos

mmm

robjohn

you have $3^{3x+1}=3^2\cdot2^{x+3}$

now we can divide both sides by $3^2$, which is the same as multiplying them by $3^{-2}$

Nicolás Castellanos

@robjohn that is we have

the left side is so 3 ^ ((3x+1) - 2)

robjohn

yes

Nicolás Castellanos

3^(3x-1)

robjohn

9:00 PM

yep

Nicolás Castellanos

the right side

conmutative property

right?

robjohn

sure

Nicolás Castellanos

3^2 / 3^2 = 1

2^(x+3) / 3^2 =

robjohn

that is true

Nicolás Castellanos

isnt the same base

robjohn

9:01 PM

but you already had $3^2$ on that side

Nicolás Castellanos

so is cancelled

and is 2^(x+3)

robjohn

so you get $2^{x+3}$ on the right

Nicolás Castellanos

i am understanding!!!

robjohn

So you have $3^{3x-1}=2^{x+3}$

Nicolás Castellanos

mmm

let me mean

or think

i dont know the word

let me use my brain

must be solved with ln

but i dont know how

robjohn

9:06 PM

Let's take the log_2 of both sides...

$(3x-1)\log_2(3)=x+3$

Nicolás Castellanos

why from this form?

$log_2\left(3^{3x-1}\right)=log_2\left(2^{x+3}\right)$

no?

robjohn

@NicolásCastellanos that brings the $x$s out from the exponent

Nicolás Castellanos

oh

robjohn

yes.

Nicolás Castellanos

and the other two?

is used to the base of the logarithms?

the base of the right side

@robjohn we have that

then..

wait

calculate the log2(3)?

no

the left side becomes

3log_2(3)x - log_2(3)

$3\log _2\left(3\right)x-\log _2\left(3\right)=x+3$

now add log_2(3) in both sides?

no

robjohn

9:14 PM

@NicolásCastellanos hang on a sec and let me look

Nicolás Castellanos

i will take a break

wait me

robjohn

$(3\log_2(3)-1)x=3+\log_2(3)$

to move all the $x$'s to one side and the rest to the other

Nicolás Castellanos

so

whats next?

the next equation is more hard :(

now divide de both sides

by 3log_2(3)-1

$\frac{\left(3\log _2\left(3\right)-1\right)x}{3\log _2\left(3\right)-1}=\frac{3}{3\log _2\left(3\right)-1}+\frac{\log _2\left(3\right)}{3\log _2\left(3\right)-1}$

yes?

right?

$x=\frac{3+\log _2\left(3\right)}{3\log _2\left(3\right)-1}$

x = 1.22