Integer solutions of log equation: log_4(X^{27}) = [log_4(X)]^4

[bobrossu]

Heres a problem I need help with:

Find all integer solutions of log₄X²⁷ = (log₄X)⁴

Possible solutions:

a) 0
b) 1
c) 1,27
d) 1,64
e) no solution

Not sure what to do, I tried to change things into exponential form. Please show me how to solve the problem with work so I can learn.

 

[Deleted member 4993]

Heres a problem I need help with:

Find all integer solutions of log₄X²⁷ = (log₄X)⁴

Possible solutions:

a) 0
b) 1
c) 1,27
d) 1,64
e) no solution

Not sure what to do, I tried to change things into exponential form. Please show me how to solve the problem with work so I can learn.

Hint:

27 = 3³

and

log₄X²⁷ = 27 * log₄X

 

[bobrossu]

Hint:

27 = 3³

and

log₄X²⁷ = 27 * log₄X

Oh! I see.

so I can turn 27 into 3³ and divide both sides by (log₄X) in order to simplify into 3³=(log₄X)³ which simplifies into log₄X=3 and X = 64

But where does the 1 come from, or is it none of the above (*not no solution,oops*)?

 

[stapel]

so I can turn 27 into 3³ and divide both sides by (log₄X) in order to simplify into 3³=(log₄X)³ which simplifies into log₄X=3 and X = 64

But where does the 1 come from, or is it none of the above (*not no solution,oops*)?

I'm not following what you're doing. Did you maybe mean something along the lines of the following?

. . . . .\(\displaystyle \log_4(x^{27})\, =\, \log_4^4(x)\)

. . . . .\(\displaystyle 27\, \cdot\, \log_4(x)\, =\, \log_4^4(x)\)

. . . . .\(\displaystyle 0\, =\, \log_4^4(x)\, -\, 27\, \cdot\, log_4(x)\)

. . . . .\(\displaystyle 0\, =\, \log_4(x)\, \left[\log_4^3(x)\, -\, 3^3\right]\)

...and then you factored the difference of cubes, set both useful factors equal to zero, etc,...?

If you meant something else, please reply showing your steps and reasoning. Thank you!

 

[bobrossu]

I'm not following what you're doing. Did you maybe mean something along the lines of the following?

. . . . .\(\displaystyle \log_4(x^{27})\, =\, \log_4^4(x)\)

. . . . .\(\displaystyle 27\, \cdot\, \log_4(x)\, =\, \log_4^4(x)\)

. . . . .\(\displaystyle 0\, =\, \log_4^4(x)\, -\, 27\, \cdot\, log_4(x)\)

. . . . .\(\displaystyle 0\, =\, \log_4(x)\, \left[\log_4^3(x)\, -\, 3^3\right]\)

...and then you factored the difference of cubes, set both useful factors equal to zero, etc,...?

If you meant something else, please reply showing your steps and reasoning. Thank you!

log₄X²⁷=(log₄X)⁴

27*log₄X=(log₄X)⁴

3³=(log₄X)³ (I divided both sides by log₄X)

3=log₄X

4³=X (exponential form)

 

[mmm4444bot]

You divided by log_4(X), without knowing its value. If you choose to divide by a symbolic number, then you have already decided that it does not equal zero.

If it turns out that a symbolic number may equal zero (that is, a zero value makes the equation true), then you will have missed the corresponding solution.

log_4(X) = 0 when x = 1

Hence, your approach divided by zero, in the case where X = 1. That's why you missed one of the solutions.

Try stapel's approach. :cool:

 

[bobrossu]

You divided by log_4(X), without knowing its value. If you choose to divide by a symbolic number, then you have already decided that it does not equal zero.

If it turns out that a symbolic number may equal zero (that is, a zero value makes the equation true), then you will have missed the corresponding solution.

log_4(X) = 0 when x = 1

Hence, your approach divided by zero, in the case where X = 1. That's why you missed one of the solutions.

Try stapel's approach. :cool:

Thanks! stapel's solution seems more logical than what I was doing.

 

[JeffM]

I'm not following what you're doing. Did you maybe mean something along the lines of the following?

. . . . .\(\displaystyle \log_4(x^{27})\, =\, \log_4^4(x)\)

. . . . .\(\displaystyle 27\, \cdot\, \log_4(x)\, =\, \log_4^4(x)\)

. . . . .\(\displaystyle 0\, =\, \log_4^4(x)\, -\, 27\, \cdot\, log_4(x)\)

. . . . .\(\displaystyle 0\, =\, \log_4(x)\, \left[\log_4^3(x)\, -\, 3^3\right]\)

...and then you factored the difference of cubes, set both useful factors equal to zero, etc,...?

If you meant something else, please reply showing your steps and reasoning. Thank you!

I do not quite understand this answer.

I am with you up through

\(\displaystyle 0 = \{\log_4(x)\}^4 - 27log_4(x) = log_4(x) * [\{log_4(x)\}^3 - 27].\)

The second term is a difference of cubes and so can be factored, but why bother? The zero product property requires that

\(\displaystyle 0 = log_4(x) \implies x = 1 \text { or}\)

\(\displaystyle 0 = \{log_4(x)\}^3 - 27 = \{log_4(x)\}^3 -3^3 \implies \{log_4(x)\}^3 = 3^3 \implies\)

\(\displaystyle log_4(x) = 3 \implies x = 4^3 = 64.\)

Very minor issue. I just wonder what nuance I am missing.

 

[mmm4444bot]

I do not quite understand this answer …

… I just wonder what nuance I am missing.

I don't think you're missing anything, Jeff. You finished one way; stapel finished a different way. Both methods lead to the same solutions.

By factoring the Difference of Cubes, I note that stapel's approach does include the extra step of realizing the quadratic factor's roots are not Real (i.e., the discriminant is negative), and, therefore, the quadratic factor can be ignored, but the remaining two factors match the ones you used. :cool:

 

[JeffM]

I don't think you're missing anything, Jeff. You finished one way; stapel finished a different way. Both methods lead to the same solutions.

By factoring the Difference of Cubes, I note that stapel's approach does include the extra step of realizing the quadratic factor's roots are not Real (i.e., the discriminant is negative), and, therefore, the quadratic factor can be ignored, but the remaining two factors match the ones you used. :cool:

Ahh thanks mmm.

Yes, if we had a polynomial more complex than

\(\displaystyle u^{(2k-1)} - a\) remaining as a factor,

we would have to factor that further. So the general rule is to factor down to linear or quadratic terms. We can take a short cut as I did when we have a term like u^3 - a because such a term has only one real zero.

EDIT: By the way, as a matter of pedagogy, this kind of problem would be great for teaching substitution of variables.

 

[JeffM]

Substitution of variables lets us see patterns. It could usefully be introduced before calculus.

\(\displaystyle log_4(x^{27}) = \{log_4(x)\}^4 \implies 27 * log_4(x) = \{ log_4(x)\}^4.\)

\(\displaystyle \text {Let } u = log_4(x).\)

\(\displaystyle \therefore 27u = u^4 \implies u^4 - 27u = 0 \implies u(u^3 - 27) = 0 \implies\)

\(\displaystyle u = 0 \text { or } u^3 = 27 \implies u = 3.\)

\(\displaystyle u = 0 \implies log_4(x) = 0 \implies x = 4^0 = 1.\)

\(\displaystyle u = 3 \implies log_4(x) = 3 \implies x = 4^3 = 64.\)

Substitution of variables does add steps, but it is very clear.

Great exercise. Involves logs, exponentials, implied division by zero, and opportunity for substitution of variables.

 

[mmm4444bot]

Great exercise. Involves logs, exponentials, implied division by zero, and opportunity for substitution of variables.

Agree!!

I would add it to my list -- if I could find the darn thing.