How Do I Solve a Natural Log for x?

[OpticDean]

Please Help! Natural Log Question!

Here is the question that is bothering me:

I need to solve this natural log for x. Please I need step by step instructions on how to figure out x. Thanks very much.

ln (x) + ln (x+1) = 2

 

[Diane_]

Remember the rules of logarithms: log(ab) = log a + log b

That's kind of the original point of logs, actually, if you remember.

So - do it backwards:

ln x + ln(x + 1) = ln(x(x+1)) = ln(x^2 + x) = 2

Does that get you far enough?

 

[Pyrrhus]

Use logarithm properties

$$\ln (AB) = \ln(A) + \ln(B)$$

 

[OpticDean]

Yes, I was able to get that far, but how do I get X all by itself on one side, like [ x = blah, blah]

thanks

 

[Diane_]

ln(a) = b implies a = e^b

 

[Pyrrhus]

$$x^2 + x = e^2$$

By e i mean antiln.

 

[OpticDean]

yes, so it becomes e^2 = x^2 + x

Now maybe my question is, what is the value of x, in order for that original equation to be true, x = ?

Thanks

 

[Pyrrhus]

Heh, Diane and myself seems to be helping almost in sync

 

[Pyrrhus]

Remember, Quadratic Equations.

$$Ax^2 + Bx + C = 0$$

$$x^2 + x + (-e^2) = 0$$

 

[OpticDean]

I finally got it, x = 2.26388...

Thank you very much Cyclovenom and Diane, you've ended my day-long struggle (its pathetic i know, but its been a while)

You guys are brilliant