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=$\pm$2
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=$\pm$2
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

 

[CellCoree]

/09/2021

Thank you for your question, Victor. When solving equations involving logarithms, it is important to remember that logarithms are only defined for positive numbers. Therefore, if the base of the logarithm is positive, the argument must also be positive. This is why in your example, when using logarithms, we only obtain the positive solution for x.

To answer your first question, when a is rational, there can be one or two solutions for x. This depends on whether the value of b is positive, negative, or zero. If b is positive, there will be two solutions (one positive and one negative). If b is negative, there will be no real solutions. If b is zero, there will be one solution (x=0).

When a is not rational, the number of solutions for x can vary greatly. It becomes a more complex problem that may require advanced mathematical techniques to solve.

When a or b is not a real number, the equation may not have any real solutions. This is because logarithms are only defined for positive real numbers. If a or b is a complex number, the equation may have complex solutions.

In summary, the number of solutions for x in an equation involving logarithms depends on the values of a and b, and whether they are real or complex numbers. It is important to keep in mind the restrictions on the values of the logarithm's argument to ensure we obtain valid solutions. I hope this helps clarify your questions.