anonymous
  • anonymous
Use the properties of logarithms to evaluate the expression Ln e^sqrt(7) I know the answer is 1, but I don't know why it's 1. Can someone explain?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
it's sqrt7
zepdrix
  • zepdrix
Here is an approach you could take. A rule of logarithms:\[\large \color{royalblue}{\log(a^b)=b\cdot \log(a)}\]
zepdrix
  • zepdrix
Understand how we can apply this rule? \[\Large \ln\left(e^{\sqrt7}\right)=?\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
So \[Ln(e^\sqrt{7}) = Log _{e}e ^{\sqrt{7}} = \sqrt{7} * Log _{e}e\] I don't quite get how this equals 1. I'm not supposed to use a calculator to understand it.
zepdrix
  • zepdrix
Ok good that's our first step. Hmm if you can't use a calculator then you need to try and remember some things about logs.|dw:1377657189784:dw|
zepdrix
  • zepdrix
|dw:1377657291950:dw|
zepdrix
  • zepdrix
This is kind of a weird way to think about it maybe :) But it's important to understand how exponentials and logs relate.
anonymous
  • anonymous
I know that works, but the problem doesn't specify the answer. It's multiple choice, and a calculator told me that answer was 1. I just don't know how it got there from \[Ln _{e}e ^{\sqrt{7}}\]
anonymous
  • anonymous
Log base e of e to the power of the square root of 7, not Ln base e. My bad.
zepdrix
  • zepdrix
Did you enter it incorrectly into your calculator I assume? :o The answer is certainly not 1 :d
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=ln+e^sqrt%287%29 Says the result is 1, which is one of the multiple choice answers. If that is wrong could you show me how to get to the correct answer?
zepdrix
  • zepdrix
http://www.wolframalpha.com/input/?i=ln+%28e%5Esqrt%287%29%29
zepdrix
  • zepdrix
Yah you're not entering it correctly. Wolfram interpreted that as the power being applied to the entire log.\[\Large \ln(e^{\sqrt7}) \ne \ln(e)^{\sqrt7}\]
zepdrix
  • zepdrix
brackets are important :c
anonymous
  • anonymous
Yes, that makes sense, but I still do not understand how exactly the answer is \[\sqrt{7}\]
zepdrix
  • zepdrix
Look back at my last picture, do you understand why the \(\Large ?\) will equal 1?
anonymous
  • anonymous
It makes sense, because the only thing that e to the power of that equals e is 1, but I can't see how that works with the problem I have.
zepdrix
  • zepdrix
Yah logs are a little tricky! :O So here is what we're doing in this problem: \[\Large \ln\left(e^{\sqrt{7}}\right)\] Applying a rule of logs let's us write it as: \[\Large \sqrt{7}\cdot\ln\left(e\right)\] Then realizing that ln(e)=1 let's us write it retricess, \[\Large \sqrt{7}\cdot\color{royalblue}{\ln\left(e\right)}\] this, \[\Large \sqrt{7}\cdot\color{royalblue}{1}\]
anonymous
  • anonymous
I think it will remain a mystery to me why Ln(e) = 1, but other than that, I understand it. Thank you for your help!
anonymous
  • anonymous
Well, I understand why 1 is the only solution to Ln(e), but it's still weird!
zepdrix
  • zepdrix
weird stuff!+_+

Looking for something else?

Not the answer you are looking for? Search for more explanations.