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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!

 Apr 25, 2018
 #1
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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 

 Apr 25, 2018

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