Solving a log equation with (3) different bases

  • Thread starter Thread starter pumaking94
  • Start date Start date
  • Tags Tags
    Bases Log
pumaking94
Messages
5
Reaction score
0
I used to think I was master of any high school math course until I came across this log equation in a calculus textbook.

<br /> \log_{2}x + \log_{5}(2x+1) = \ln x<br />

I use the change of base to convert everything to one base and then I get down to:

<br /> (2x+1)^{\ln2} = x^{\ln2\ln5 - \ln5}<br />

I have no idea how to solve that. I found a website where they use graphing and later say that the algebraic solution is too difficult for beginner students. So what is the algebraic solution?

PS Sorry for posting in "General Math" but this isn't really a homework question...
 
Last edited:
Physics news on Phys.org
Welcome to PF pumaking94! :smile:

I'm afraid there is no algebraic solution using only a finite number of standard functions.
So there's no such thing as being too difficult. It just isn't there. :wink:
 
Thanks!

That is disappointing, this site kept saying "The algebraic solution is too difficult" so I thought there definitely has to be one ;p, but I guess not.

What do you mean by finite number of standard functions? What about an infinite number?
 
There are methods to write an expression as an infinite series.
Perhaps you already know of some of those.
For instance:
\frac 1 {1-x} = 1 + x + x^2 + x^3 +...
Note that this is only true for |x| < 1.

(This may not be a very good example, since the original expression can be written with standard operations.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top