Integrated Circus - 2021-04-08

robjohn

20:01

room topic changed to Integrated Circus: 3 rings without identity (no tags)

Nicolás Castellanos

alright

robjohn

in Mathematics, 50 mins ago, by Nicolás Castellanos

$8^{-3x}\cdot 2^{x+1}=4^{x+2}$

Nicolás Castellanos

we have $-2^{-8x+4}x+1=-2^{2x+7}x+1$

robjohn

The Math room was getting agitated.

Nicolás Castellanos

yes

robjohn

20:04

@NicolásCastellanos no you don't have that

Nicolás Castellanos

you told me that multiply the two sides by -8x+1

lets start from the beginning

robjohn

$2^{-8x+1}=2^{2x+4}$ is what we are starting with

I suggested multiplying both sides by $2^{8x-1}$

Nicolás Castellanos

ok

robjohn

$2^{-8x+1}\cdot2^{8x-1}=2^{2x+4}\cdot2^{8x-1}$

Nicolás Castellanos

the left side is: $-2^{-8x+4}x+1$

why

$2^{-8x+1}\cdot -8x+1$

=

robjohn

20:07

@NicolásCastellanos NO

Nicolás Castellanos

ah

robjohn

ADD the exponents

Nicolás Castellanos

you told me multiplying the sides by 8x+1, and that confused me

wait

robjohn

$2^0=2^{10x+3}$

@NicolásCastellanos that is NOT what I said

Nicolás Castellanos

in the other chat

well lets forget it

robjohn

20:09

3 mins ago, by robjohn

I suggested multiplying both sides by $2^{8x-1}$

Nicolás Castellanos

:(

robjohn

in Mathematics, 26 mins ago, by robjohn

Why not multiply both sides by $2^{8x-1}$?

Nicolás Castellanos

the left side: 2^(-16x-2)?

robjohn

the sum of the exponents on the left side is $0$

Nicolás Castellanos

why?

robjohn

20:11

$(-8x+1)+(8x-1)$

Nicolás Castellanos

ah

im reading bad

the right side

2^(10x + 5

robjohn

$(2x+4)+(8x-1)=???$

Nicolás Castellanos

10x+3

robjohn

yes

Nicolás Castellanos

2^0 = 1

2^(10x+3) = 1

10x+3 = 0

robjohn

20:13

$10x+3=0$

yes

Nicolás Castellanos

and now clean the expression

robjohn

$x=-\frac3{10}$

Nicolás Castellanos

10 to divide

robjohn

yep

Nicolás Castellanos

x = -0.3

robjohn

20:15

yes

Nicolás Castellanos

that equation already has the "= 0"

i can solve it joining the two 3 bases?

or the form is cuadratic formula?

i will make it my self

i have this

$x=\frac{\left(-3+-\sqrt{153}\right)}{6}$

correct?

robjohn

That is a quadratic equation, so you would use the formula

but what you get is $3^x=\frac{-1\pm\sqrt{1^2-4\cdot(-12)}}{2}$

if $y=3^x$, the equation is then $y^2+y-12=0$ ($a=1,b=1,c=-12$)

So...

Nicolás Castellanos

20:33

im sorry i was busy

but a = 3, b = 3, and c = -12

simplify the equation?

nono 3^x

robjohn

no

Nicolás Castellanos

i can understand

why 3^x and no x?

robjohn

You are solving $y^2+y-12=0$ and $y=3^x$

Nicolás Castellanos

oh

y^2 = 3^2x

robjohn

I think what you mean is correct, but the way you've written it has problems. $y^2=3^{2x}$

which would be y^2=3^(2x)

Nicolás Castellanos

20:37

ok

robjohn

otherwise people might think you mean y^2=9x

but yes, $y^2=3^{2x}=9^x$

Nicolás Castellanos

$y=\frac{\left(-1+-\sqrt{49}\right)}{2}$

sqrt(49) = 7

with +, y = 3

3^x = 3

x = 1

robjohn

yes

Nicolás Castellanos

$3^2+3^1-12 = 0$

now with minus

robjohn

You won't be able to get a negative number for $3^x$ with real $x$, so you can ignore the other root from the quadratic equation

Nicolás Castellanos

20:42

y=-4

robjohn

yes...

Nicolás Castellanos

someone in the math chat told me that i need the two solutions

robjohn

@NicolásCastellanos not with real numbers. With complex numbers you could have many solutions

So whoever said you needed two was not correct.

$3^x=-4$ has no real solutions

Nicolás Castellanos

yes

PM 2Ring

@NicolásCastellanos I said:

in Mathematics, 4 hours ago, by PM 2Ring

@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

20:46

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

20:51

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

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

20:54

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

20:56

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

21:00

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

21:01

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

21:06

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

21:14

@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

too hard

simplify

$1.21^{\left(5x+5\right)}+1.1^{\left(10x+10\right)}=1.1$?

now what?

@robjohn ???

mick

21:45

hi

Nicolás Castellanos

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PM 2Ring

22:04

@NicolásCastellanos $121=11^2$, so $1.21=1.1^2$

robjohn

22:14

OMG this room has been invaded!

@NicolásCastellanos that is correct

Yes. $1.21=1.1^2$

So apply the quadratic equation again

which is greater, $1.21^{5(x+1)}$ or $1.1^{10(x+1)}$?

I take it back. You don't need the quadratic equation

PM 2Ring

22:57

@NicolásCastellanos $log(1) = 0$

Nicolás Castellanos

ok

whats the relation of a with b if log(a) + log(b) = 0

that means that log(ab) = 0

that means that ab = 1

that means that 1 / a = b

PM 2Ring

@NicolásCastellanos Correct

Transcript for

$Mathematics$ Integrated Circus