Logarithmic Bases and Exponential Functions
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When you DIVIDED by loge(x)
You had assumed that:\(\displaystyle log_e(x) \ \ne \ 0 \)
Thus you DISCARDED one of the solutions.
Whenever you divide, you need to be always aware of these "unspoken" assumption/s.
However, excellent job in "questioning" methods!!
That assumption that loge(x) ≠ 0 has to be true though?
If it was 0, then it would be 0^2 / 0 which is NOT equal to 1...
HallsofIvy
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No, you are making the same mistake again, dividing by ln(x). If ln(x) were 0 you couldn't divide by ln(x)= 0 at all! If x= 1 so that ln(x)= 0, the equation becomes 0= 0 which is a true equation.
No, you are making the same mistake again, dividing by ln(x)
^ But Subhotosh Khan said "when you DIVIDED by loge(x)" so I thought it was from that step that was the issue?
f ln(x) were 0 you couldn't divide by ln(x)= 0 at all!
^ Why can't you divide by 0?
If x= 1 so that ln(x)= 0, the equation becomes 0= 0 which is a true equation.
^ If x =1 then that makes it 0^2/0 = 1 which 0 = 1 is not true...
The natural number system has a useful property, namely there is a smallest number.
6 - 7 is not a natural number.
You can create a number system, namely the integers, where 6 - 7 is a number. But then you lose that important property that there is a smallest number.
Can you create a number system where division of a non-zero real number by zero is a number? Yes, you can. But then we are not dealing with the real number system at all. We lose certain properties of the real numbers. In one of the ways to create a number system that allows (with a single exception) division by zero, we lose for example the distinction between smaller and bigger. Or to put it another way, - 3 is both bigger and smaller than 3. Most mathematicians do not think the cost of permitting division by zero is worth the gain.
If we want to have the attributes of number that are present in the real number system, division by zero cannot be not a defined operation. Permitting it would lead to bizarre and highly undesirable results.
[MATH]a = b \implies a^2 = ab \implies a^2 - b^2 = ab - b^2 \implies [/MATH]
[MATH](a - b)(a + b) = b(a - b) \implies a + b = b \implies a + a = a \implies 2a = a \implies 2 = 1.[/MATH]
Now that result is clearly wrong; it is the result of dividing by (a - b), which is zero because a = b.
So, unless you want learn a whole different kind of math, never divide by zero. If you do, your results will be wrong.
Oh I should have rephrased my question differently... I meant why can't you divide by 0 in the context that it would be 0/0 which would just be 0
If loge(x) = 0, by changing to exponential form you get:
e^0 = x
hence, x = 1 is a solution.
How are you getting 0^2/0 = 1?
A division by zero is not defined. If a division by 0 is made then problems may arise.
Simply put if 1/0 = x,
then multiplying both sides by 0 would give:
1=0
which is clearly, as you stated before, incorrect.
Now because loge(x) can actually be zero, the problem that arises when you divide by it is that you lose a solution. This is why your method is invalid and does not provide all solutions.
However, this does not mean that your method does not work. In some cases, e.g. 1/x, there is a natural restriction or there can be a restriction in the question that allows you to assume that a variable (x) CANNOT be 0 and thus a division by x is valid.
Even in the one-point compactification of the reals, 0/0 is indeterminate.
[MATH]5 * 0 = 0 \implies 5 = \dfrac{0}{0}.[/MATH]
[MATH]4 * 0 = 0 \implies 4 = \dfrac{0}{0}.[/MATH]
[MATH]\therefore 4 = 5.[/MATH]
YOU CANNOT DIVIDE ZERO INTO ANY REAL NUMBER AND GET A UNIQUE REAL RESULT.
[MATH]5 * 0 = 0 \implies 5 = \dfrac{0}{0}.[/MATH]
[MATH]4 * 0 = 0 \implies 4 = \dfrac{0}{0}.[/MATH]
[MATH]\therefore 4 = 5.[/MATH]
YOU CANNOT DIVIDE ZERO INTO ANY REAL NUMBER AND GET A UNIQUE REAL RESULT.
Steven G
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10/2 = 5, since only 2*5=10
20/5 = 4, since only 5*4=20
0/0 = 0, since 0*0 = 0
0/0 = 23, since 0*23 = 0
0/0 = sqrt(5), since 0*sqrt(5) = 0
Which is it, does 0/0 equal 0, 23 or sqrt(5)?
0/0 is indeterminate since we can't decide on what it should equal.
0/0 certainly is NOT 0
HallsofIvy
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Writing that a= b/c is the same as writing ac= b. We cannot write a= b/0 (for b non-zero) because that is the same as writing a(0)= b which is not true for any a. We cannot write a= 0/0 because that is the same as a(0)= 0 which is true for all numbers, not just some specific a.
Lots of confusion from reading all of the comments, however I THINK I might understand now:
My understanding from this is that when I divided the left-hand-side by loge(x), I (unknowingly) assumed that loge(x) is not equal to 0, otherwise the division cannot be done due to the fact that 0/0 is indeterminate (which I did not know until now!). This is why I should not use division in my method of solving the question, rather I should just use the first method in my attachment.
Is this correct???
My understanding from this is that when I divided the left-hand-side by loge(x), I (unknowingly) assumed that loge(x) is not equal to 0, otherwise the division cannot be done due to the fact that 0/0 is indeterminate (which I did not know until now!). This is why I should not use division in my method of solving the question, rather I should just use the first method in my attachment.
Is this correct???
Technically its because any division by 0 is undefined. But yes, you are correct.
A simple example to show this is
x2 - x = 0
If you divide both sides by x (invalid) you get:
x - 1 = 0
x = 1 (but a quadratic can have 2 solutions)
Hence why we factorise (valid):
x(x-1) = 0 and apply the null - factor theroem
x-1 = 0
x = 1
or: x = 0
I’d rather say that when you seem to need to divide by a function of a variable, you always need to consider the cases where the value of the function is zero as special cases and solve those special cases without division. So I would not say not to use division. Just consider the special cases separately and avoid division there.
EDIT: In the previous post, the example was given of
[MATH]x^2 - x = 0 \implies x^2 = x.[/MATH]
I want to divide by x, the identity function. But wait; what if the value is zero.
[MATH]x = 0 \implies x^2 = 0 \implies x^2 = x \ { if } x = 0.[/MATH]
[MATH]x \ne 0 \text { and } x^2 = x \implies \dfrac{x^2}{x} = \dfrac{x}{x} \implies x = 1.[/MATH]
The factoring method is far more efficient for this problem. But the special cases method is much more general.
EDIT: In the previous post, the example was given of
[MATH]x^2 - x = 0 \implies x^2 = x.[/MATH]
I want to divide by x, the identity function. But wait; what if the value is zero.
[MATH]x = 0 \implies x^2 = 0 \implies x^2 = x \ { if } x = 0.[/MATH]
[MATH]x \ne 0 \text { and } x^2 = x \implies \dfrac{x^2}{x} = \dfrac{x}{x} \implies x = 1.[/MATH]
The factoring method is far more efficient for this problem. But the special cases method is much more general.
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