How Does Logarithmic Solution Limit the Roots of x^a=b?

[n_kelthuzad]

if x^a=b (a,b are constants)
then there are two ways of finding x: root and log
so for example, x^2=4
by root:
(x^2)^(1/2)=(4)^(1/2)
x=\pm2
by log:
2 ln (x) = 2 ln 2
x=2
but it is yet impossible to obtain the negative x from logs. How are you supposed to do it? And here's a few questions:
1.when a is rational how do you know x have 1 or 2 answers?
2.what happens when a is not rational?
3.what happens when a or(and) b is not real?


thanks
Victor Lu
16

 

[maxbur]

x^a=b
log[x](b)=a <--- [x] is base x
log(b)/log(x)=a
(1/a)*log(b)=log(x)
log(b^(1/a))=log(x)
b^(1/a)=x

 

[maxbur]

also i think i can say that there is no difference in whether you change a to rational or irrational, and there will always be two solutions, as it will always be some root
but i don't know about non-real numbers, sorry.

 

[arildno]

if x^a=b (a,b are constants)
then there are two ways of finding x: root and log
so for example, x^2=4
by root:
(x^2)^(1/2)=(4)^(1/2)
x=\pm2
by log:
2 ln (x) = 2 ln 2
x=2
but it is yet impossible to obtain the negative x from logs. How are you supposed to do it? And here's a few questions:
1.when a is rational how do you know x have 1 or 2 answers?
2.what happens when a is not rational?
3.what happens when a or(and) b is not real?


thanks
Victor Lu
16

Hi, Victor!

when you take the SQUARE root of x^2, the CORRECT answer is root(x^2)=|x|, not x.
|x| is what we call "the absolute value" of the number x, i.e, its distance from 0 (irrespective of direction), which is always a non-negative number.

Thus, solving x^2=4 with the square root operation gives you FIRST:
|x|=2

Then, when you wish to remove the absolute value sign, you get two solutions.

root(4)=2 always, never -2