Question

solve ln2+lnx^2=3

Answer 1

Hi :)
Recall this log rule: \(\large\rm \log(a)+\log(b)=\log(ab)\)
We can apply this to the left side of our equation.

Answer 2

\(\large\rm \ln(2)+\ln(x^2)=ln(?)\)

Answer 3

=ln2x^2

Answer 4

@zepdrix

Answer 5

yes that is right.
So you have,

\(\color{#000000 }{ \displaystyle \ln\left(2x^2\right)=3 }\)

Answer 6

Then, there is another rule you need to know:

\(\color{#000000 }{ \displaystyle a=\log_bz\quad \quad \Longrightarrow \quad \quad b^a=z }\)

And since \(\ln\) is same as \(\log_e\).
Therefore,

\(\color{#000000 }{ \displaystyle a=\ln z=\log_ez\quad \quad \Longrightarrow \quad \quad e^a=z }\)

Answer 7

so i would use log to solve this

Answer 8

\(\color{#000000 }{ \displaystyle 3=\ln (2x^2) \quad \quad \Longrightarrow \quad \quad 2x^2=e^3 }\)

Answer 9

by the same rule.

Answer 10

then solve for x.

Answer 11

x=3.16

Answer 12

well, first, there are two answers. Also, can you please write the answers exactly?

Answer 13

i took \[\frac{ e \frac{ 3 }{ 3 } }{ 2\frac{ 1 }{ 2 } }\]

Answer 14

Sorry my notification didn't show up :[ weird..

Answer 15

yeah, happens to me 40 out of 55 times

Answer 16

\(\color{#000000 }{ \displaystyle 2x^2=e^3 }\)

\(\color{#000000 }{ \displaystyle x^2=e^3/2 }\)

\(\color{#000000 }{ \displaystyle \sqrt{~x^2}=\sqrt{~\frac{e^3}{2}} }\)

and then, knowing the definition of the absolute value;
\(|z|=\sqrt{z^2}\), continue please.

Answer 17

Example about this definition of absolute value?

Answer 18

so the square root will go away on the x^2

Answer 19

not exactly,

Answer 20

rather, \(\sqrt{x^2}\) will turn into \(|x|\) ("absolute value of x")

Answer 21

leaving you with \[x ^{2}=\sqrt{\frac{ e ^{3} }{ 2}}\]

Answer 22

No no... ok I will give you an example of the application of this defintion of \(\sqrt{z^2}=|z|\)

Answer 23

anything absolute cant be negative

Answer 24

but \(x^2\) ?

Answer 25

its just x

Answer 26

the square root of x^2 is x

Answer 27

No!

Answer 28

unless x>0

Answer 29

Suppose I am solving the following equation for \(t\).
\(\color{#000000 }{ \displaystyle t^2-16=0 }\)

adding 16 to both sides,
\(\color{#000000 }{ \displaystyle t^2=16 }\)

then I take the square root of both sides,
\(\color{#000000 }{ \displaystyle \sqrt{t^2}=\sqrt{16} }\)

applying the definition of absolute value,
\(\color{#000000 }{ \displaystyle |t|=4}\)

and therefore,
\(\color{#000000 }{ \displaystyle t=\pm 4}\)

So my two solutions would be:
\(\color{#000000 }{ \displaystyle t=4}\) and \(\color{#000000 }{ \displaystyle t=-4}\)

Answer 30

\(\color{#000000 }{ \displaystyle 2x^2=e^3 }\)

\(\color{#000000 }{ \displaystyle x^2=\frac{e^3}{2} }\)

\(\color{#000000 }{ \displaystyle \sqrt{~x^2}=\sqrt{~\frac{e^3}{2}} }\)

Answer 31

then, please proceed and keep in mind the definition of the absolute value.

Answer 32

so its \[\frac{ e \sqrt{2e} }{ 2 }\]

Answer 33

How did you get that result?

Answer 34

you are missing the other solution

Answer 35

i solved it with my calculator

Answer 36

yes, but there is another solution, and if you follow the example I did then this should be clear

Answer 37

i dont understand y there would be 2 answers

Answer 38

yes i am looking at it now

Answer 39

ok, tell me if you get it or not after you are done reading it

Answer 40

is the other answer 0 or something or like plus or minus the answer i got?

Answer 41

is the other answer 0 or something or like plus or minus the answer i got?

Answer 42

After squaring both sides, via the definition of the absolute value you get

\(|x|=\sqrt{\dfrac{e^3}{2}}\)

Answer 43

and you know that if |x|=a, then x=\(\pm\)a, right?

Answer 44

and same here, except that in your case, the "a" is that big square root.

Answer 45

okay

Answer 46

proceed please, and tell me the two answers you get.

Answer 47

is it \[\pm \frac{ e \sqrt{2e} }{ 2 }\]

Answer 48

yes, exactly!

Answer 49

Great

Answer 50

this is all I needed, you pleased me \(\lim_{x\to\infty}(jk)^{x} \)

Answer 51

no just kidding::)

Answer 52

lol omg

Answer 53

good luck

Answer 54

thanks for all of the help

Answer 55

i have 1 last question

Answer 56

it looks really simple but im not getting it its give the general solution for the following equation: \[\sin ^2x+2=3sinx\]

Answer 57

@SolomonZelman

Answer 58

\(\color{#000000 }{ \displaystyle \sin^2x+2=3\sin x }\)

you can substitute
\(\color{#000000 }{ \displaystyle \sin x=z }\)

and this will get you
\(\color{#000000 }{ \displaystyle z^2+2=3z }\)

Solve for z, and after you solve for z, you can solve for x knowing that \(z=sin(x)\).

Answer 59

z is 2,1

Answer 60

\(\color{#000000 }{ \displaystyle z^2-3z+2=0 }\)
\(\color{#000000 }{ \displaystyle (z-2)(z-1)=0 }\)
\(\color{#000000 }{ \displaystyle z=1,~2 }\)

Godd!

Answer 61

-1 < sin( quantity ) < 1

for whatever quantity is!

Answer 62

because sine is the ratio of the opposite side (o) and hypotenuse (h), and

|o/h | \(\le\) 1
Therefore
| sin( anything) | \(\le\) 1

Answer 63

you had 2 solutions:
z=2
z=1

Answer 64

z=2 is not a solution however

Answer 65

because if z=2, then sin(x)=2

Answer 66

And as I explained
sin(x)=2 is false for any real (or complex) x.

Answer 67

So, the only case you have left is
sin(x)=1

Answer 68

there is an infinite number of solutions for this, actually, unless x is on some interval.

Answer 69

can you give some solutions to this equation tho', please?

Answer 70

so sin is 2

Answer 71

The sin is to miss-follow my comments. No just kidding.

I have however explained that sin(x)=1 is the only case, although I admit I might have done that to briefly and quickly.

Answer 72

sin( of what? ) = 0

can you give me some examples, please?

Answer 73

(you may refer to the unit circle too, if you like)

Answer 74

lol well pi is

Answer 75

Sure, good!

sin(pi)=0

Answer 76

sin(0)=0

Answer 77

i have a unit circle chart

Answer 78

I will write this:
\(\color{#000000 }{ \displaystyle \sin(x)=0\quad\Longrightarrow \quad x=k\times \pi;~\forall k\in\mathbb{Z} }\)

Answer 79

can you read that notation?

Answer 80

it's fine if you can't

Answer 81

I will translate

Answer 82

"if sin(x)=0, then x=k•π, for any integer k"

Answer 83

So,

x=-1000π
x=-23π
x=0π=0
x=9623π
x=2π
x=49π
x=-127π

and so on, for any integer in front of π, that is a solution

Answer 84

Why?

Answer 85

A π is a circle, correct?

Answer 86

So, knowing that when you subtract π from 0, or add π to 0, you return to the same point where y=0.

Answer 87

So, you can add or subtract any number of π, as long as you subtract or add whole number of π's - i.e. go as many full circles as you like in either clockwise or anticlockwise direction, you will be returning to the same point on the unit circle.

Answer 88

yes i understand that

Answer 89

Well, that is the general concept why you can add pi or subtract pi as many times as you want (as long as you add pos/neg integer amount of pi's.

Answer 90

so whenever you have
\(\sin[f(x)]=0\)

And you get 1 solution, say
\(x=r\)

then,
\(k\times \pi+r\)

would be the solutions for all integer k

Answer 91

kπ is just the number of whole circles you travel, but you always, again, return to same point.

Answer 92

Well, \(\sin[f(x)]=t\)

doesn't have to =0

Answer 93

anyways, won't overwhelm with this, since I myself am not very good with unit circle.

Answer 94

good luck!

Answer 95

lol you seem pretty good with it to me are you like a math professor or something