Find the domain of the functions?

Question

a) f(x)=log(base 2)(logx)

b) f(x)=log(base 3)(9-x^2)

 

How to get an answer of:

a)x>1

b) -3<x<3

Answer 1

f(x) = log₂ (log x).

log x > 0   No power (log₂) results in a negative.

If x = 1, then log x = 0. This requires log₂ to produce 0, which it cannot.

Therefore x > 1.

 

The same reasoning applies to b).

 

Answer 2

Edward,

 

I will show you how to solve a related a problem, which should help you solve your two problems.

 

Domain of log₆(3+x)?

 

log₆(x) will return an a that makes 6^a = x. So log₆(36) returns 2, since 6² = 36.

 

Since there is no a that will make 6^a = 0 or a negative number, x must be ≥ 0. log₆(-1) DNE, since 6^a is always positive. log⁶(0) doesn't exist for the same reason.

 

 

If you don't believe me, try raising 6 to some numbers and see if the result's zero or negative. Try 6^-100, 6^-1, 6⁰, 6¹, 6¹⁰, etc. Try graphing 6^a.

 

ANSWER: log₆(3+x) will only accept arguments that are greater than 0. So domain log₆(3+x) is 3+x>0 → x>-3.

 

 

HINTS

 

Problem 1: log₂(log(x)) = log₂(log₁₀(x))

 

The argument (the thing inside the parenthesis) of log₂() must be greater than 0 for reasons described above. The same is true for the argument of log₁₀(). Think about the doman of log₁₀(x). Then think about its range, part of which is necessarily log₂(log₁₀(x))'s domain.

 

Plug in numbers (x=-1, 0, 1, 10, 100) if you're not sure.

 

 

Problem 2: log₃(9-x²)

 

This one is actually easier. The argument of log₃() must be greater than 0. That means 9-x2 > 0.

 

Good luck and happy Saturday,

 

Bryce