NicholausBlackmon
  • NicholausBlackmon
Can someone help me figure out how to solve this question? In 4 + In (3x)=2
Mathematics
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
NicholausBlackmon
  • NicholausBlackmon
Note: I have absolutely no idea how to. I completely forgot
SolomonZelman
  • SolomonZelman
You need to know: \(\color{#000000 }{ \displaystyle \ln(a)+\ln(b)=\ln(a\cdot b) }\) \(\color{#000000 }{ \displaystyle \ln(z) =y\quad \Longrightarrow \quad z=e^y }\)
SolomonZelman
  • SolomonZelman
Use the first rule to combine the logarithms on the left side.

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NicholausBlackmon
  • NicholausBlackmon
can you explain how i should plug them in because like i said, I have absolutely no idea. This was given to me as bonus for my teacher last week and she said that i could still do it
SolomonZelman
  • SolomonZelman
In your case, a is 4 b is 3x
SolomonZelman
  • SolomonZelman
If you know; \(\color{#000000 }{ \displaystyle \ln(a)+\ln(b)=\ln(a\cdot b) }\) Then, \(\color{#000000 }{ \displaystyle \ln(4)+\ln(3x)=? }\)
NicholausBlackmon
  • NicholausBlackmon
so its in (4) + in(3x)=2?
SolomonZelman
  • SolomonZelman
"in" ?
SolomonZelman
  • SolomonZelman
That is ell, not i. "ln" denotes/abbreviates - natural logarithm
NicholausBlackmon
  • NicholausBlackmon
ln. sorry
SolomonZelman
  • SolomonZelman
Can you apply the [first] rule that I provided?
NicholausBlackmon
  • NicholausBlackmon
Im on my brothers account btw. Im in 7th grade and have absolutely no idea what ln means. Can you please explain it to me?
SolomonZelman
  • SolomonZelman
have you heard of a function \(e^x\) ?
NicholausBlackmon
  • NicholausBlackmon
yes, i have
SolomonZelman
  • SolomonZelman
Ok, very good
SolomonZelman
  • SolomonZelman
|dw:1450329327758:dw|
SolomonZelman
  • SolomonZelman
That is approximately what this function looks like
NicholausBlackmon
  • NicholausBlackmon
okay
SolomonZelman
  • SolomonZelman
time out for short.... do you know how to write \(e^{-10}\) with a positive exponent?
NicholausBlackmon
  • NicholausBlackmon
yes i do
SolomonZelman
  • SolomonZelman
ok, go ahead please
SolomonZelman
  • SolomonZelman
The rule is: \(a^{-b}=1/a^b\)
NicholausBlackmon
  • NicholausBlackmon
is it 4.5?
SolomonZelman
  • SolomonZelman
no
SolomonZelman
  • SolomonZelman
you need to give me an exact value for \(e^{-10}=1/e^{10}\) (this is what I wanted) Ok, can you do \(e^{-100}\) ?
NicholausBlackmon
  • NicholausBlackmon
ohhhhh
SolomonZelman
  • SolomonZelman
\(e^{-100}\) please(?)
NicholausBlackmon
  • NicholausBlackmon
give me a second please
SolomonZelman
  • SolomonZelman
sure
NicholausBlackmon
  • NicholausBlackmon
1/e^100?
SolomonZelman
  • SolomonZelman
Yes
SolomonZelman
  • SolomonZelman
(I am going over some preliminary things)
SolomonZelman
  • SolomonZelman
And \(1/e^{100}\) is smaller than \(1/e^{10}\). Because \(e^{100}\) is bigger than \(e^{10}\) and that means that in \(1/e^{100}\) you are dividing 1 by a bigger number and thus the result is smaller.
NicholausBlackmon
  • NicholausBlackmon
okay
SolomonZelman
  • SolomonZelman
And so is \(\color{#000000 }{ \displaystyle y=e^{x}}\) For negative x-values, the more negative you go, the more the output (y) is going to approach zero.
NicholausBlackmon
  • NicholausBlackmon
that makes sense
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f(x)=e^{x}}\) \(\color{#000000 }{ \displaystyle f(-10)=e^{-10}=1/e^{10} }\) \(\color{#000000 }{ \displaystyle f(-100)=e^{-100}=1/e^{100} }\) \(\color{#000000 }{ \displaystyle f(-1000)=e^{-1000}=1/e^{1000} }\) \(\color{#000000 }{ \displaystyle f(-10000)=e^{-10000}=1/e^{10000} }\) see the value of the output is going to zero (but never actually hits zero)
SolomonZelman
  • SolomonZelman
And obviously, \(\color{#000000 }{ \displaystyle f(x)=e^{x} }\) will get infinitely large for larger values of x...
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f(20)=e^{20} }\) \(\color{#000000 }{ \displaystyle f(400)=e^{400} }\) \(\color{#000000 }{ \displaystyle f(6000)=e^{6000} }\) and it gets bigger
SolomonZelman
  • SolomonZelman
|dw:1450330076981:dw|
NicholausBlackmon
  • NicholausBlackmon
so f(8000) is greater than f(6000)?
SolomonZelman
  • SolomonZelman
yes (in case f(x)=e^x)
SolomonZelman
  • SolomonZelman
very good
NicholausBlackmon
  • NicholausBlackmon
oh okay
SolomonZelman
  • SolomonZelman
and the bigger x we choose the bigger the f(x) gets...
NicholausBlackmon
  • NicholausBlackmon
yea i see now
SolomonZelman
  • SolomonZelman
So, again, this is the graph of \(e^x\) |dw:1450330210762:dw| and the \(y=\ln(x)\) would be...
SolomonZelman
  • SolomonZelman
|dw:1450330231177:dw|
NicholausBlackmon
  • NicholausBlackmon
My brother says the answer to the equation is 1.41. Is he correct?
NicholausBlackmon
  • NicholausBlackmon
But hes not making any sense on how he got the answer
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle \ln(4\cdot 3x)=2 }\) \(\color{#000000 }{ \displaystyle \ln(12x)=2 }\) \(\color{#000000 }{ \displaystyle e^2=12x }\) \(\color{#000000 }{ \displaystyle e^2/12=x }\)
SolomonZelman
  • SolomonZelman
even if you say that e=3 (but e<3), then you stll get less than 1, so the result 1.41=x, is not right
NicholausBlackmon
  • NicholausBlackmon
oh
NicholausBlackmon
  • NicholausBlackmon
okay so the answer has to be less then 1 on this equation
SolomonZelman
  • SolomonZelman
Very good observation!
NicholausBlackmon
  • NicholausBlackmon
sorry for pointing out the obvious
SolomonZelman
  • SolomonZelman
no, no it is fine :)
SolomonZelman
  • SolomonZelman
you can get more precise on the reasoning, e<3 Therefore e^2/12 < (3)^2/12 e^2/12 < 9/12 e^2/12 < 3/4 the result is less than 3/4
SolomonZelman
  • SolomonZelman
it is about 0.6157 (there is a by hand technique for approximating ln 's and e^x's but this is calculus I at least)
NicholausBlackmon
  • NicholausBlackmon
now he says its . 62
SolomonZelman
  • SolomonZelman
62 ?
SolomonZelman
  • SolomonZelman
He probably has no clue about what ln(x) even means -: (
NicholausBlackmon
  • NicholausBlackmon
.62
SolomonZelman
  • SolomonZelman
oh, that is correct
SolomonZelman
  • SolomonZelman
to two dec. places.
NicholausBlackmon
  • NicholausBlackmon
thank you! I really do appreciate your help!
SolomonZelman
  • SolomonZelman
Well, I kind of rather blurred out the answer than introduced anything new.
NicholausBlackmon
  • NicholausBlackmon
its fine, hes trying to show me now since he finally got the correct answer
SolomonZelman
  • SolomonZelman
but if you know what inverse function is, then \(\ln(x)\) and \(e^x\) are inverses of each other.
SolomonZelman
  • SolomonZelman
They got corresponding behaviors, but reversed... I will show the graphs and what I am talking about.
NicholausBlackmon
  • NicholausBlackmon
okay
SolomonZelman
  • SolomonZelman
|dw:1450330755906:dw||dw:1450330782267:dw|
NicholausBlackmon
  • NicholausBlackmon
i see
SolomonZelman
  • SolomonZelman
Look at the graphs, isn't ln(x) kind of e^x spinned ?
SolomonZelman
  • SolomonZelman
In fact, they are inverses and reflect each other about the origin.
SolomonZelman
  • SolomonZelman
they "mirror" about the line y=x.
NicholausBlackmon
  • NicholausBlackmon
Oh wait. we can still type
NicholausBlackmon
  • NicholausBlackmon
please dismiss that message i sent you
SolomonZelman
  • SolomonZelman
|dw:1450330944025:dw|
NicholausBlackmon
  • NicholausBlackmon
okay, i see
SolomonZelman
  • SolomonZelman
And \(\ln(x)\) has some properties such as addition of logarithms \(\color{#000000 }{ \displaystyle \ln(a)+\ln(b)=\ln(a\cdot b) }\) difference of logarithms \(\color{#000000 }{ \displaystyle \ln(a)-\ln(b)=\ln(a\div b) }\) COMPARE TO: \(e^x\) has some properties: \(\color{#000000 }{ \displaystyle e^{a+b}=e^a\cdot e^b }\) \(\color{#000000 }{ \displaystyle e^{a-b}=e^a\div e^b }\) (something to think about)
SolomonZelman
  • SolomonZelman
I am probably not going to be able to teac the entire logarithm course just via this not too awesome drawing tool, and without face-to-face.... kind of hard that way...
SolomonZelman
  • SolomonZelman
one good source: http://www.mathsisfun.com/algebra/exponents-logarithms.html
NicholausBlackmon
  • NicholausBlackmon
Its okay, i understand
SolomonZelman
  • SolomonZelman
good luck :)
NicholausBlackmon
  • NicholausBlackmon
thanks!
SolomonZelman
  • SolomonZelman
anytime

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