Logarithm equation with different base

In summary: The equation may be able to be solved numerically, but that's a pretty intensive process that wouldn't be feasible for most people.
  • #1
songoku
2,292
325

Homework Statement


Find x that satisfies:

log11 (x3 + x2 - 20x) = log5 (x3 + x2 - 20x)


Homework Equations


logarithm


The Attempt at a Solution


x3 + x2 - 20x = 1

x3 + x2 - 20x - 1 = 0

Then stuck...
 
Physics news on Phys.org
  • #2
Edit: Disregard that. I was incorrect.
 
Last edited:
  • #3
songoku said:

Homework Statement


Find x that satisfies:

log11 (x3 + x2 - 20x) = log5 (x3 + x2 - 20x)

Homework Equations


logarithm

The Attempt at a Solution


x3 + x2 - 20x = 1

x3 + x2 - 20x - 1 = 0

Then stuck...
Actually folks, this is correct:
x3 + x2 - 20x = 1 .​
The only way for log5(u) = log11(u) is for u = 1 .

That cubic function does not factor nicely.
 
  • #4
bossman27 said:
I'm not sure what you're trying to do in your attempt at a solution...

Think of it this way:

[itex] log_{11}(x^{3}+x^{2}-20 x) = log_{5}(x^{3}+x^{2}-20x) [/itex]

means [itex] 11^{k} = 5^{k} = x^{3}+x^{2}-20x [/itex].

So what can "k" be for this: [itex] (11^{k} = 5^{k}) [/itex] to hold? I think you might have assumed it must be 1, but that is not correct.

Since RHS and LHS have different base, the only possible case for RHS = LHS is when the argument of the log = 1

For your hint:
11k = 5k is only correct for k = 0. It means that the argument of the log = 0, which is not correct since log 0 is undefined.

Or maybe I get it wrong?

Thanks
 
  • #5
songoku said:
x3 + x2 - 20x - 1 = 0

Then stuck...

You are not lucky with your book of problems. That equation has only complex roots. Check it at wolframalpha.com.
http://www.wolframalpha.com/input/?i=x^3+%2B+x^2+-+20x+-+1+%3D+0

ehild
 
  • #6
ehild said:
You are not lucky with your book of problems. That equation has only complex roots. Check it at wolframalpha.com.
http://www.wolframalpha.com/input/?i=x^3+%2B+x^2+-+20x+-+1+%3D+0

ehild

Hm...using my calculator, I got three real roots. They are:

x = - 4.9776
x = 4.02750
x = - 0.04988

And those numbers are really similar to the x-intercept of root plot (diagram) of wolframalpha. But I don't understand why there are also three alternatives complex solutions.

OK, the point is this equation can't be solved manually.

Thanks a lot for the help :smile:
 
  • #7
songoku said:
Hm...using my calculator, I got three real roots. They are:

x = - 4.9776
x = 4.02750
x = - 0.04988

And those numbers are really similar to the x-intercept of root plot (diagram) of wolframalpha. But I don't understand why there are also three alternatives complex solutions.

OK, the point is this equation can't be solved manually.

Thanks a lot for the help :smile:

Well, even wolframalpha can be wrong.

ehild
 
  • #8
songoku said:
Hm...using my calculator, I got three real roots. They are:

x = - 4.9776
x = 4.02750
x = - 0.04988

And those numbers are really similar to the x-intercept of root plot (diagram) of wolframalpha. But I don't understand why there are also three alternatives complex solutions.

OK, the point is this equation can't be solved manually.

Thanks a lot for the help :smile:
The fact that the roots given by WolframAlpha are complex, is likely an artifact of whatever numerical method is being used there. The imaginary part of the roots listed there is extremely small relative to the real part, with the real part matching the roots as you have listed them and the imaginary part on the order of 10-15 .
 

What is a logarithm equation with different base?

A logarithm equation with different base is an equation that involves a logarithm function with a base other than the typical base 10 or base e. The equation is written as logb(x) = y, where b is the base, x is the argument, and y is the result.

What is the difference between a logarithm equation with different base and a standard logarithm equation?

The main difference between a logarithm equation with different base and a standard logarithm equation (log10(x) or ln(x)) is the base used. In a standard logarithm equation, the base is either 10 or e, whereas in a logarithm equation with different base, the base can be any positive number other than 1.

How do you solve a logarithm equation with different base?

To solve a logarithm equation with different base, you can use the change of base formula: logb(x) = log(x) / log(b). This formula allows you to convert the equation into a standard logarithm equation, which can then be solved using basic algebraic techniques.

What are the common applications of logarithm equations with different base?

Logarithm equations with different base have many applications in fields such as finance, computer science, and physics. They are commonly used in financial modeling, calculating exponential growth, and measuring the performance of algorithms. They are also used in physics to model radioactive decay and in chemistry to calculate pH levels.

What are the properties of logarithm equations with different base?

The properties of logarithm equations with different base are similar to those of standard logarithm equations. Some of the key properties include the product rule, quotient rule, and power rule. These properties allow you to simplify and solve complex logarithm equations with ease.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
3K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
9K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
5
Views
993
  • Precalculus Mathematics Homework Help
Replies
2
Views
707
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
Back
Top