Solving a log equation with (3) different bases

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Homework Help Overview

The discussion revolves around a logarithmic equation involving different bases, specifically: \(\log_{2}x + \log_{5}(2x+1) = \ln x\). The original poster expresses difficulty in solving the equation after attempting to use the change of base formula.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve the equation by converting it to a single base and simplifies it to an expression involving powers. They inquire about the existence of an algebraic solution. Other participants discuss the limitations of algebraic solutions and introduce the concept of infinite series.

Discussion Status

The conversation is ongoing, with participants exploring the nature of solutions to the equation. Some guidance has been offered regarding the absence of a finite algebraic solution, and the possibility of expressing the equation in terms of infinite series has been mentioned.

Contextual Notes

There is a mention of the original poster's concern about the complexity of the algebraic solution and the context of the problem being from a calculus textbook. The discussion also touches on the distinction between finite and infinite functions in relation to solving the equation.

pumaking94
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I used to think I was master of any high school math course until I came across this log equation in a calculus textbook.

[itex] \log_{2}x + \log_{5}(2x+1) = \ln x[/itex]

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

[itex] (2x+1)^{\ln2} = x^{\ln2\ln5 - \ln5}[/itex]

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...
 
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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:
[tex]\frac 1 {1-x} = 1 + x + x^2 + x^3 +...[/tex]
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.)
 

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