Question

solve ln(x-2)-ln(x+1)=lnx

Answer 1

Would you like help on this?

Answer 2

yes please

Answer 3

Alright cool! So do you have an idea of what we should do first?

Answer 4

would you have to equal the (x-2) and (x+1) to each other ?

Answer 5

unfortunately they are logs so we can't really do so. But we can take advantage of the fact that they are in the ln's. Do you know how to get rid of the lns?

Answer 6

no :c im actually extremely lost

Answer 7

haha so you need to put both sides as the exponents of e, does that make sense?

Answer 8

like so:
\[\ln(x-2)-\ln(x+1) = lnx\]\[e^{\ln(x-2)-\ln(x+1)} =e^{ lnx}\] does this seem familiar?

Answer 9

no :C ive actually never done a question like this one. my teacher posted this up as extra credit and i would appreciate it sooo much if you guided me step by step with explanation on how to do it ..

Answer 10

ahhh nice extra credit! That's always a good thing. Sure, but to do this problem you will need to learn some things, is that ok with you?

Answer 11

yeah i have no problem, i am in ap calculus so soon he will bring this question into topic. i am honestly not very good in calc. i end up forgetting everything the next day :( and thats my issue. but yes, i am so for it teach me everything

Answer 12

oooh calc, that was a long time ago but it was enjoyable. Are you doing AB or BC?

Answer 13

AB

Answer 14

Anyways sorry for getting off topic...ok so you need to know this identity:
\[10^{\log(x)} = x = log(10^{x})\] does this make sense?

Answer 15

kind of .. yeah

Answer 16

alright so just keep that in your mind btw ln is just that but instead of 10 it's for e, does that make sense?

Answer 17

oh okay so ln replaces the 10 with e ? and e and 10 are the same ??

Answer 18

no e is called the natural base, it's a really important number in the world. ln is the natural logarithm.

Answer 19

if you have a scientific or graphing calculator it should have that value on it

Answer 20

ooh okay

Answer 21

ok so to show you how that works:
\[e^{lnx} = x\] make sense?

Answer 22

okay

Answer 23

so \[e^{ln(x-2)-ln(x+1)}= e^{ln(x)}\] simplifies down to
so \[e^{ln(x-2)-ln(x+1)}= x\] does that makes sense?

Answer 24

oooh okay yes got it, continue.

Answer 25

alrighty now we need to know about another rule, so here it is:
\[x^{a-b} = \frac{x^{a}}{x^{b}}\] have you seen this before?

Answer 26

no i have not.

Answer 27

Does it make sense though?

Answer 28

yes :)

Answer 29

alright cool so then our:\[e^{\ln(x-2)-\ln(x+1)} = x\] becomes this:\[\frac{e^{\ln(x-2)}}{e^{\ln(x+1)}} = x\] make sense?

Answer 30

yes

Answer 31

but then it simplifies down further because we got the whole e^ln thing:
\[\frac{x-2}{x+1} = x\] do you know how to solve it now?

Answer 32

no :(

Answer 33

well we then move the (x+1) to the other side so it becomes:
\[x-2 = x(x+1)\] which is \[x-2 = x^{2} + x\] and it becomes \[-2 = x^{2}\] which we can solve to be \[x= \pm i \sqrt{2}\] does that make sense?

Answer 34

yes

Answer 35

Cool and that's the problem! So not that bad right?

Answer 36

no its not :o easier than i thought it would be. once you presented me with the formulas i was able to understand how they connected. i also need help in one more question, if you dont mind.

Answer 37

uhh sure, I might be a little slow in responding tho if that's alright with you?

Answer 38

yeah its cool :)
\[(1000)^x+2(100)^x-3(10)=0\]

Answer 39

I am assuming you have no idea how to start this?

Answer 40

nope .. :x

Answer 41

haha ok so do you notice something peculiar with how they gave you the question?

Answer 42

it decreases ? going fromm 1000 to 100 to 10 and equaling to 0 ?

Answer 43

haha well there's that but aren't 1000, 100, and 10 all powers of 10?

Answer 44

haha im just laughing at myself

Answer 45

yes you are right haha

Answer 46

so why don't we put everything as a power of 10, do you know how to do so?

Answer 47

ummm replacing the x with 10 ??

Answer 48

if i replace x with 10 the answer would be a really long number

Answer 49

no all you do is this:
\[(1000)^{x}+2(100)^{x}-3(10)=0\]
\[10^{3^{x}}+ 2(10^{2^{x}})-3(10) = 0\] make sense?

Answer 50

oh i see

Answer 51

now all we do is the same thing we did to the previous problem but instead of making everything the power of e, we simply log both sides and use the identity I showed you previously. Does this make sense?

Answer 52

so you log both sides ? and which identity ?

Answer 53

the whole:
\[x^{a-b}=\frac{x^{a}}{x^{b}}\]

Answer 54

part do you see it?

Answer 55

oooh okay so i log both sides and then plug it into that ? or ..

Answer 56

well that's how you separate the whole mess you'll have when you log both sides...oh I see, I'm sorry, that identity is the wrong one...uhhh let see...

Answer 57

its okay

Answer 58

uhh let me think about it for a little bit, I am like helping multiple people so I get confused every once in awhile...

Answer 59

its okay haha but was that identity incorrect in general ?

Answer 60

no it's a correct identity you just would use it in this case lol.

Answer 61

not

Answer 62

is what I mean...

Answer 63

ok I got...sorry about that!

Answer 64

let's start over ok?

Answer 65

its okay :)

Answer 66

alright so we start off with this right?
(1000)^x+2(100)^x - 3(10) = 0
what we want to do is actually is divide both sides by 10:
(100)^x + 2(10)^x-3 = 0 does this make sense?

Answer 67

yes

Answer 68

alright now we can just say that z = 10^x, do you understand we can just arbitrarily do this?

Answer 69

yes

Answer 70

ok cool so if you use that we can substitute our previous equation thusly:
x^2 + 2x -3 = 0, do you get how I got this? from this: 100^x+2(10^x)-3 = 0?

Answer 71

yes :)

Answer 72

cool and now we can just solve that quadratic right? so what is the solution to that quadratic?

Answer 73

(x+3)(x-1)=0 ?

Answer 74

yup so x = -3,1. But wait!!! That's not x right? that's actually z! so z = -3,1 but we want x, so we go back to how we defined z: z = 10^x or x = log(z)...Therefore x = log(-3),log(1). But wait! we can't have a log of a negative number of log(-3) doesn't make any sense. And log(1) is equal to 10. So now we have our final solution of x = 10! Does this all make sense?

Answer 75

where did z come from though ?

Answer 76

remember we arbitrarily defined previous that z = 10^x? Like a few comments ago.

Answer 77

oooh yes yes sorry

Answer 78

so does everything make sense?

Answer 79

yes :)

Answer 80

nice! sorry for leading you astray! But I'm glad that you understood everything!

Answer 81

thank you so much !! have a good night :)

Answer 82

you as well!