hey guys i got my exam tomorrow and i cant nail this question
my teacher wrote the answer = x=3/4 but that makes no sense to me ...
thanks in advance
hey mate,
for any equation of the form
f(x)g(x) = 0
either f(x) = 0, g(x) = 0 or f(x), g(x) = 0
as f(x) = 3^x, it theoretically (and I am cringing as I write this) never can be 0 even as x approaches -infinity and as such for your situation the only solution can lie when g(x) = 3^x - 27 = 0
thus,
3^x = 27
your teachers solution is incorrect as the only solution for this is x = 3 as 3^3 = 27 (this can easily be verified with a basic logarithmic analysis). Are you a high school student, if so what country are you residing in??? if Australian (as I am) never trust a 'maths' teachers word as gospel, the requirements to teach maths at a secondary level is pathetic.
This should be your classroom
Hope this helps,
David
Hey Mr Fantastic,
thats what I thought, just wanted to make sure, just out of interest, have you encountered a proof that any real number raised to a complex number can never be zero??? and moreover how can one evaluate a real number a raised to the complex value i ???
Cheers,
David
I'm assuming from your response that the direct way of addressing
a^(i) is to simply use Euler's Identity
e^(ix) = cos(x) + isin(x)
thus as any real positive value a can be expressed as e^ln(a), i.e.
a = e^(ln(a)
then,
a^(i) = (e^ln(a))^ i = e^(i * ln(a)) = cos(ln(a)) + i*sin(ln(a))
Thus for,
a^x = 0, where x can be complex, a is real and positive the expression becomes,
let,
x = c + di
then
a^x = 0 --> a^(c + di) = 0
--> a^c * a^(di) = 0
as c, d are real a^c ~= 0 thus we are left with
a^(di) = 0 or (a^d)^i = 0
as a, d are real, let a^d = f (nohing a^d is positive)
hence we obtain,
f^i = 0
from before f^i = cos(ln(f)) + isin(ln(f)) = 0
equating reals and imaginaries we have
cos(ln(f)) = 0 , sin(ln(f)) = 0
Thus, ln(f) = (2n+1)*pi/2 , ln(f) = n*pi , n being an integer
So for a solution to exist to a^x = 0 then there must be an integer n such that,
(2n+1)*pi/2 = n*pi,
or,
n*pi =pi/2 which has no real solutions for n being an integer.
would you consider this to be a sufficient 'proof' and I say that word losely for why a^z = 0 is a is real, and if your in agreeance, what are you thoughts if a is complex???
Regards,
David