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 :)

Printable View

- Nov 20th 2008, 01:26 AMjimzerHelp 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 ...

thanks in advance :) - Nov 20th 2008, 02:19 AMmr fantastic
- Nov 20th 2008, 02:21 AMjimzer
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?

- Nov 20th 2008, 02:27 AMmr fantastic
- Nov 20th 2008, 02:37 AMDavid24
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 - Nov 20th 2008, 02:40 AMjimzer
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 - Nov 20th 2008, 02:40 AMDavid24
- Nov 20th 2008, 02:46 AMmr fantastic
- Nov 20th 2008, 02:50 AMDavid24
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 - Nov 20th 2008, 02:52 AMmr fantastic
- Nov 20th 2008, 03:17 AMDavid24
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