negative logarithms and the multiplication law?

Question

ok so im a bit confused about logarithm laws.
i will put the logarithm base in [ ]

the math book im studying has an example:
log[8](2-x)+log[8](4-x)=1
it works out to be x=0 and x=6
the next part of the question asks me to check these answers by substitution.

when x=0 it works out fine.

when x=6 it works out to log[8](-4)+log[8](-2)=1

at this point my math book says "it is impossible to find an exponent to base 8 that will produce -4. same applies to finding an exponent to base 8 that will produce -2" it then tells me therefore that the left hand of this equation is undefined.

my question is this, can you not use the law of logarithms for multiplication?
my book has it as log[a]xy = log[a]x + log[a]y a>0, a cannot = 1

doing this, it would become
log[8](-4)+log[8](-2)=1
log[8](-4)(-2)=1
log[8]8=1
log[8]8^1=1
1=1

it makes sense to me to do it this way. am i wrong? or is the book wrong?

Answer 1

For your syllabus of studies, log[a] b is "undefined" for when b is negative values. Probably because your syllabus haven't reach the topic regarding Imaginary Number and Natural Number.



As you pursue for higher studies, like in college maths, you may encounter Euler's identity which proves that e^(iπ) = -1.

e = natural number ; i = sqrt(-1)



When taking Log of base natural number {written as: Log[e] or ln}, the equation becomes:

Log[e] (-1) = iπ



Take the example in your case:

Log[8] (-4) = Log[e](-4) / Log[e](8)

Log[8] (-4) = {Log[e](-1) + Log[e](4)}/ Log[e](8)

Log[8] (-4) = {iπ + Log[e](4)} / Log[e](8)

Check this out: http://www.wolframalpha.com/input/?i=Log8%28-4%29



Similarly, Log[8] (-2) = {iπ + Log[e](2)} / Log[e](8)



Therefore

Log[8](-4) + Log[8](-2) = {iπ + Log[e](4)} / Log[e](8) + {iπ + Log[e](2)} / Log[e](8)

Log[8](-4) + Log[8](-2) = {Log[e](4) + Log[e](2) + 2iπ} / Log[e](8)

Log[8](-4) + Log[8](-2) = {Log[e](8) + 2iπ} / Log[e](8)

Log[8](-4) + Log[8](-2) = 1 + 2iπ/Log[e](8)





Hope I'm not sound confusing, good luck!





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EDIT:



Log[8](-4)(-2)=1 ....is Mathematically correct.



But you can't make deduction that

Log[8](-4)+log[8](-2) = log[8](-4)(-2)

(...which is Mathematically incorrect!)



This is because in the first expression

Log[8](-4)+log[8](-2)

tells you that there is imaginary number involved (and cannot be ignored).



'Jumping' the expression to

Log[8](-4)(-2)

would neglect the imaginary number in the first expression.





By the way, just as in all Logarithmic rules and identities;

Log[a] (xy) = Log[a](x) + Log[a](y)

only valid for a>0, x>0, y>0 ....as wrote on your textbook, remember?!

Answer 2

You are slightly wrong. While what you did is 99% correct, it is possible to perform seemingly valid operations on an equation that result in a loss of accuracy.



For example, if presented with the equation f(x) = (x^2 - 4)/(x + 2) you might reduce it to f(x) = x - 2, but in doing so you have made a tiny change in the equation - the original equation looked just like f(x) = x - 2, but was undefined at x = -2, and if you had looked inifinitely closely at the line at that point, there would be a break. It is a tiny difference, but real, and in properly reducing it you must state it like f(x) = x - 2 (for x <> -2) to maintain that difference.



Likewise, in the logarithm you showed the original equation is undefined at certain points, and while you can perform mathematical operations that seem to remove that issue, they really don't. This is a very important part of applying mathematics to real life - sometimes your original equation will contain these domain issues that you must maintain, or applying your solution could be disastrous. Likewise, sometimes an equation will give multiple possible solutions of which some might not be physically applicable (like imaginary results or negative lengths.).

Answer 3

Hmm... you make an interesting point. I suppose the argument against it will be that each piece of the logarithm must be valid, so your approach is incorrect. On the other hand, when dealing with imaginary numbers, we can multiply two imaginary numbers to get a real number, e.g. 2i * 3i = -6.



Maybe you can be the inventor of a new class of numbers. Call them "imaginary logarithms". :)



* * * * *



Cosmicfish: Yeh, but I don't see where the poster's logic violates anything. He is just using the rule that log a + log b = log (ab). What's the problem with that?



We see quirky things like this quite a lot in mathematics, actually. For example, two functions f(x) and g(x) can be discontinuous at point a, but f(x) + g(x) can be continuous at a - in other words, two values that don't exist somehow combine to form a value that exists. This can also happen with imaginary numbers.



Why not do something similar with logarithms? log(-3) and log(-2) do not exist, but I don't see why we can't say log(-3) + log(-2) exists. If a logical inconsistency can't be demonstrated, then it seems to me this is not much different from saying that √-3 * √-2 is a real number.

Answer 4

the logs can be written as log8[ (2-x)(4-x) ] = 1

antilog base 8 gives

(2-x)(4-x)= 8

which has solutions x = 0 and x = 6

but as you say, 2-6 = -6 and log8(-4) is undefined

therefore, since x = 0 does not have these "issues" only x = 0 is a feasible solution

Answer 5

a) x = log (base 3) of (a million/eighty one) you're able to try this on a scientific calculator: take the log of a million/eighty one and divide with the aid of the log of three, and you get -4 yet you additionally can try this on your head. evaluate this: 3^2 = 9 9^2 = eighty one 3^4 = eighty one 3^(-4) = a million/eighty one hence, x = -4 b) on condition that log(a) + log(b) = log(ab), hence: log (base 2) of x + log (base 2) of (x + 2) - 2 = log (base 2) of (x(x + 2)) - 2 = log (base 2) of (x^2 + 2x) - 2