Solving a logarithmic equation

Logarithms problems.

Solving a logarithmic equation

Postby tenzilkeli3 » Fri Oct 02, 2020 10:33 am

Hey all, I've gotten stuck on a question while studying for alg II final, and was hoping that somebody may be able to help (teacher is not available until late next week).

The equation is this: log( x3 )=x-94

For the life of me I cannot figure out how to solve it. I have tried manipulating it in different ways using the algebraic properties of logarithms to no avail. I feel like I'm missing something obvious, and I need some help seeing it!

Thanks in advance!
tenzilkeli3
 

Re: Solving a logarithmic equation

Postby Guest » Mon Oct 05, 2020 5:16 pm

An equation like this, where the unknown, x, is both "inside" and "outside" the logarithm (or expontial), cannot be solved with "elementary" methods. You will have to get a numerical solution or use some variation of "Lambert's W function" (defined as the inverse function to [tex]f(x)= xe^x[/tex]). I can't imagine such a problem being given in an "Algebra II" class!
Guest
 

Re: Solving a logarithmic equation

Postby Guest » Sun Jan 16, 2022 8:35 pm

Since this has been here a while:
(I am assuming the "x3" is "x^3" and that the logarithm is the natural logarithm)
log(x^3)= 3 log(x)= x- 94
log(x)= x/3- 94/3
x= e^(x/3- 94/3)= e^(x/3)e^(-94/3)
xe^(-x/3)= e^(-94/3)

Let y= -x/3 so x= -3y
-3ye^y= e^(-94/3)
ye^y= -e^(-94/3)/3

Now apply "Lambert's W function" (also called the "product log function") to both sides. Since it is the inverse function to ye^y we get y= -x/3= W(-e^(-94/3)/3)

x= -3W(e^(-94/3)/3).

I doubt you would see such a problem in a secondary school class.
Guest
 


Return to Logarithms, Log, ln



Who is online

Users browsing this forum: No registered users and 1 guest