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dxs's avatar

What happened to the solutions of this logarithmic equation?

Asked by dxs (14404points) 6 days ago

Our math class is talking about logarithms now, and we came across a really interesting problem while solving a logarithmic equation: 2^^(log2(x^^2))=4, log2 means log of base 2, m^^n means m to the nth power.

Here’s how I solved it:

1. 2^^(log2(x^^2)) = 4

2. log2(4) = log2(x^^2)

3. 4 = x^^2

4. x = +/- 2

Here’s how the other person solved it:

1. 2^^(log2(x^^2)) = 4

2. 2^^(log2(x^^2)) = 2^^2

3. log2(x^^2)= 2

4. 2 log2(x)=2

5. log2(x) = 1

6. x=2

One of the solutions vanished! We thought it had something to with the property used in line 3 of her proof: log(X^^c) = c log(X). log(X^^c) = 0 has c solutions by the Fundamental Theorem of Algebra, while c log(x) = 0 only requires one. This property seems to deplete the number of solutions. How can one justify this phenomenon? It doesn’t seem mathematically sound to me. Where did the solutions go?

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6 Answers

RedDeerGuy1's avatar

They decided not to use imaginary or complex numbers. Like the square root of + and – 2

zenvelo's avatar

I think you lose it in Line 2 of her solution. it could be 2^^2, or (-2)^^2.

4 has both a positive square root and a negative square root.

Mariah's avatar

Pretty sure @zenvelo has it. I’ll look more closely tomorrow.

LostInParadise's avatar

The problem occurs in going from line 3 to line 4. log2(x^^2) = 2 log2(x) only holds when x is positive.

For example, log2((-2)^^2) = 2
Misapplying the rule gives 2 log2(-2) = 2 and then log2(-2)= 1, which is not true.

dxs's avatar

@zenvelo I don’t think that matters:

1. 2^^(log2(x^^2)) = (-2)^^2
2. 2^^(log2(x^^2)) = (-1*2)^^2
3. 2^^(log2(x^^2)) = ((-1)^^2*2^^2)
4. 2^^(log2(x^^2)) = 2^^2

Say c*log2(X) = log2(X^^c) = 1. Then c*log2(X) = 1 and log2(X^^c)=1. c*log(X) = 1 means 2^^1/c=x, or x is the c^th root of 2, which can have c solutions. log2(X^^c)=1 means 2=x^^c, or x^^c – 2 = 0, which has c solutions.

But, as @LostInParadise, pointed out, (-2)^^x is crazy. How many solutions does (-2)^^x -1 = 0 have? Does it interfere with the Fundamental Theorem of Algebra? I mean, I see that it appears not to be true in this case, but perhaps in a more general sense why? I’ll think more about this.

P.S. Sorry for the sloppy notation. To add to it, Fluther does strange things if you have only one caret.

LostInParadise's avatar

When you talk about logarithms of negative numbers then you have to use complex numbers for the logarithms and things get messy. There is no contradiction with the Fundamental Theorem of Algebra, because that theorem only talks about polynomials, not raising to variable exponents.

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