Logarithm equation with different base

[songoku]

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...

 

[bossman27]

Edit: Disregard that. I was incorrect.

 

[SammyS]

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.

 

[songoku]

I'm not sure what you're trying to do in your attempt at a solution...

Think of it this way:

log_{11}(x^{3}+x^{2}-20 x) = log_{5}(x^{3}+x^{2}-20x)

means 11^{k} = 5^{k} = x^{3}+x^{2}-20x.

So what can "k" be for this: (11^{k} = 5^{k}) 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:
11^k = 5^k 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

 

[ehild]

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

 

[songoku]

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

 

[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

Well, even wolframalpha can be wrong.

ehild

 

[SammyS]

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

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 .