# Help please ... Simple Exponential Function

• November 20th 2008, 12:26 AM
jimzer
Help please ... Simple Exponential Function
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 ...

• November 20th 2008, 01:19 AM
mr fantastic
Quote:

Originally Posted by jimzer
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 ...

Either $3^x = 0$ (impossible for real values of x) or $3^x - 27 = 0 \Rightarrow 3^x = 27 = 3^3 \Rightarrow x = 3$.
• November 20th 2008, 01:21 AM
jimzer
thnaks a lot!! is it because you can divide both sides by 3^x to get rid of the 3^x? and the 0 would still remain 0?
• November 20th 2008, 01:27 AM
mr fantastic
Quote:

Originally Posted by jimzer
thnaks a lot!! is it because you can divide both sides by 3^x to get rid of the 3^x? and the 0 would still remain 0?

It's because if AB = 0 then one or both of A and B have to equal zero.
• November 20th 2008, 01:37 AM
David24
Quote:

Originally Posted by jimzer
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 ...

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.

Hope this helps,

David
• November 20th 2008, 01:40 AM
jimzer
yeah true

yeah dude im in hi school

i live in noble park, go to st bedes in mentone

and your right itd only work if the equation iwas 3^x(3^x) - 27 = 0 for x to equal 3/4

thanks a lot
• November 20th 2008, 01:40 AM
David24
Quote:

Originally Posted by mr fantastic
It's because if AB = 0 then one or both of A and B have to equal zero.

hey mate,

just out of curiousity, if x wax complex, what values (and or set of solutions) would satisfy the equation??

Cheers,

David
• November 20th 2008, 01:46 AM
mr fantastic
Quote:

Originally Posted by David24
hey mate,

just out of curiousity, if x wax complex, what values (and or set of solutions) would satisfy the equation??

Cheers,

David

None actually, because z = 0 is a singularity of the complex logarithm function.

$3^x = 0$ is impossible for complex values of x too.
• November 20th 2008, 01:50 AM
David24
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
• November 20th 2008, 01:52 AM
mr fantastic
Quote:

Originally Posted by David24
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

Like $\left(e^{\pi}\right)^i \, ....$
• November 20th 2008, 02:17 AM
David24

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