+0  
 
0
81
1
avatar

I'm doing some math test prep for Math 126, but I'm confused as to how \(\frac{-3}{1-e^4}\) is the answer?

The Problem:

ln (x+3) - ln x = 4

 

I tried to condense the left side so that I'd have \(ln(\frac{x+3}{x})=4\) and then raise both sides to the power of \(e\), but that doesn't seem to work.

 

I thought that I could use this condensing and power raising technique since I was able to get the correct answer for the problem right after with this method, which by the way was: \({log}_{2}x+{log}_{2}(x+6)=4\) 

 

Is there something different I must do with natural log equations or was I on the right track?

 

Help would be greatly appreciated, thanks!

Guest Apr 25, 2018
 #1
avatar+87333 
+1

ln ( x + 3)  - ln x  = 4

 

ln [ (x + 3) / x ]  = 4

 

e^ [ ln [ (x + 3 ) / x ] ]  = e^4        { remember that e^(ln a)  = a  }

 

[ x + 3 ] / x  = e^4      cross-multiply

 

x + 3  = xe^4   rearrange as

 

x - xe^4  =  - 3       factor out  x

 

x [ 1 - e^4]  = -3 

 

x  = - 3 / [ 1 - e^4]   

 

 

cool cool cool 

CPhill  Apr 25, 2018

12 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.