Question

Topic: Base e and Natural Logarithms
ln (x-2) = 2
and another one is
ln x + ln 2x = 2 @whpalmer4

Answer 1

Okay, are you supposed to solve for \(x\)?

Answer 2

\[\ln x = a\]means that \[e^a = x\]
So if we want to solve my equation for \(x\), we raise \(e\) to the power of both sides:
\[e^{\ln x} = e^a\]but \[b^{\log_b x} = b\] so the left side simplifies:
\[e^{\ln x} = e^a\]\[x = e^a\]

Answer 3

With that knowledge, you should be able to solve the first problem. Do so and I'll check your answer.

Answer 4

so x-2=e^2

Answer 5

Yes, although I would write \[x = 2+e^2\] if you are asked to solve for \(x\)

Answer 6

oh ok I forgot about the 2

Answer 7

Now for the second one, you're going to use a handy property of logarithms:

\[\log a + \log b = \log (a*b)\]

Answer 8

wait what button do I press on my calculator for e

Answer 9

I would just leave the answer as you have it unless specifically asked for a decimal number.

You may have an \(e^x\) button on your calculator which calculates \(e^x\) when you've entered \(x\) already...

what model calculator?

Answer 10

\[e\approx 2.7182818284590452354\]

Answer 11

TI-84 plus

Answer 12

but I got 2.6931

Answer 13

looks like it might be the second function on the \(\ln\) button

Answer 14

oh ok I got it I have to press 2nd then the button

Answer 15

Yeah, 0.6931 is the natural log of 2. One of the values I've memorized :-)

Answer 16

so 9.3890 is the answer

Answer 17

it turns out that you can memorize just a handful of values and estimate many from that.
for example, \(\ln 8 = \ln 2^3 = 3\ln 2 = 3*0.693 \approx 2.1\) (true value is 2.07944...)

Yes, 9.38906... is your value of \(x\)

Answer 18

I see there are multiple ways they can be put :)

Answer 19

but as I suggested, as long as you know how to convert it to a number, it's generally more useful to keep it in the "symbolic" (and exact) form \(x = 2+e^2\)

Answer 20

You can always turn it into a number, but recognizing that 9.38906... is \(2 + e^2\) is much more challenging :-)

Answer 21

it is but if I do get those numbers down it would have to be when I'm not busy lol

Answer 22

would the second problem be ln 2x^2

Answer 23

My point is that in general, keeping things in forms like \(35\pi, e^2\), etc. allows you to have more information about what you are dealing with, and possibly recognize paths to a solution which you wouldn't if you just had a number.

Yes, the second problem becomes \[\ln 2x^2 = 2\]Can you solve that for \(x\)?

Answer 24

I'm not really sure what to do :(

Answer 25

Well, again, raise e to both sides:

\[e^{\ln 2x^2} = e^2\]Now simplify that the same way you did with the first one

Answer 26

wait again where did you get the e from and why di you put it in. Was it to cancel out the ln?

Answer 27

e is the base of the natural logarithm

Answer 28

\[\ln x = \log_e x\]

Answer 29

it's such a useful concept that they invented that shorthand...

Answer 30

you haven't had enough math yet to see it, but it turns out that once you get into calculus and beyond, \(e\) pops up everywhere you look...there's even a book called "E: the story of a number"

Answer 31

lol really that's fuuny

Answer 32

now I'm at 2x^2+ 7.38905....

Answer 33

There's an elegant identity that links \(e\) and \(\pi\) and \(i = \sqrt{-1}\) and \(1\) and \(0\) together:

\[e^{i\pi} +1 = 0\]

Answer 34

what is that formula called?

Answer 35

Please, you shouldn't have any decimal numbers in sight while working these problems...

\[e^{\ln 2x^2} = e^2\]\[2x^2 = e^2\]Now continue...

Answer 36

https://en.wikipedia.org/wiki/Euler's_identity

Answer 37

this might sound dumb but what do I do with the 2x^2

Answer 38

Okay, you're trying to get the left side to look like \{x = \]right?

Answer 39

ummhmmm

Answer 40

sorry, \[x=\]

Answer 41

but you have \[2x^2 = \]

Answer 42

do I divide

Answer 43

why don't you start by dividing both sides by 2?

Answer 44

oh ok lol

Answer 45

then take the square root of both sides

Answer 46

"brain to forehead: look out forehead, here comes palm!" :-)

Answer 47

lol never heard that one before, so now I;kmkn

Answer 48

sorry something is going on with my backspace button

Answer 49

another good book: "Dr. Euler's Fabulous Formula: Cures Many Mathematical Ills" :-)

Answer 50

yeah, I sometimes get OpenStudy in a state where the backspace key doesn't work. annoying!

Answer 51

I well any how how would I squareroot 2e^2

Answer 52

Your still there right

Answer 53

Yes.

\[2x^2 =e^2\]Divide both sides by 2 \[\frac{2x^2}{2} = \frac{e^2}{2}\]\[x^2=\frac{e^2}{2}\]
Take square root of both sides

Answer 54

????? I'm dead

Answer 55

Why? What is the square root of \(x^2\)?

Answer 56

x lol I got that one

Answer 57

oh so its 2

Answer 58

what is 2?

Answer 59

the answer right

Answer 60

I knew when I asked that you were going to say that :-(

Answer 61

really lol y?

Answer 62

but I was hoping that you'd tell me something else, like "that's the number of problems I have left" because it is not the answer to this problem

Answer 63

lol omg I'm never going to get this

Answer 64

\[x^2 = \frac{e^2}{2}\]\[\sqrt{x^2} = \sqrt{\frac{e^2}{2}} = \frac{\sqrt{e^2}}{\sqrt{2}}\]right?

Answer 65

then simplifying (assuming that \(x>0\)):
\[x = \frac{e}{\sqrt{2}}\]

Answer 66

actually, no need to assume, we can see that \(x > 0\) here. but in general, you can't just simplify \(\sqrt{x^2}\rightarrow x\)unless you know that \(x>0\)

Answer 67

ok I get it bu tdo you mind helping me with another problem plzzzzz

Answer 68

How could I possibly refuse? :-)

Answer 69

OMG thanx ok here we goo>>>
ln (x+3) = 4

Answer 70

@whpalmer4

Answer 71

:-(

I'm happy to help, but I do ask that you try to recognize problems of a sort we've already worked out...

Answer 72

ok lol so you want me to do this one on my own? :)

Answer 73

I think you can, yes. Happy to check the result for you...

Answer 74

Ever heard of the surgeon's motto? "See one, do one, teach one" :-)

Answer 75

ok hopefully I can DO ONE lol

Answer 76

kk first you take e and square it to ln in which they cancel. Then you take the e and square it to 4

Answer 77

51.5982 would be the answer rounded am I correct

Answer 78

could we have that in plain mathese, please? :-)

That's the correct (approximate) number. But really, I'd prefer \[x = e^4-3\]which is exact.

Answer 79

ok then lol :p

Answer 80

you think you could help me with even more problems

Answer 81

\(e\) is irrational, so it is not possible to write down the exact string of digits which represents it, because it goes on forever...

Answer 82

but our teacher wants us too so Idk

Answer 83

\(\pi\) is also irrational (meaning it can't be represented as the ratio of two integers)

Oh, if your teacher wants a decimal, then by all means, that's what you should provide. I thought I said "unless the problem asks for a decimal" when I first expressed my preference for the symbolic, exact form....

I can't be held responsible if your teacher prefers inexact answers that take up more space :-)

Answer 84

lol I guess and r u ready for properties of logs or do I HAVE YOU TIRE
D YET

Answer 85

TIRED

Answer 86

let's do another, if you're up for it...

Answer 87

but why don't you post it in a fresh question...

Answer 88

@whpalmer4
Use log\[\log_{2} 3 \approx 1.5850 and \log_{2} 5\approx 2.3219 \to approximate the value of each expression.\]

Answer 89

two reasons: 1) you can only give a single medal per question, and 2) when I flip back to this page, I have to scroll to the bottom each time, which is getting successively more tedious :-)

Answer 90

1) \[\log_{2} 25\]

Answer 91

oh ok hold then lol