The domain of a function is the set of all allowed input values. The range is the set of all possible output
values. Exponential and logarithmic functions have different restrictions.
1. Exponential functions
A transformed exponential function can be written as:
\[
f(x)=A\cdot b^{k(x-h)}+v
\]
For an exponential function, the exponent can use any real value of \(x\). Therefore, the domain is:
\[
\text{Domain}=\mathbb{R}
\]
In interval notation:
\[
(-\infty,\infty)
\]
2. Range of exponential functions
The expression \(b^{k(x-h)}\) is always positive when \(b>0\) and \(b\neq1\).
The vertical shift \(v\) creates a horizontal asymptote:
\[
y=v
\]
If \(A>0\), the exponential graph lies above the asymptote:
\[
\text{Range: } y>v
\]
If \(A<0\), the exponential graph lies below the asymptote:
\[
\text{Range: } y
If \(A=0\), the function is constant:
\[
f(x)=v
\]
In that special case, the range is:
\[
\{v\}
\]
3. Logarithmic functions
A transformed logarithmic function can be written as:
\[
f(x)=A\log_b(k(x-h))+v
\]
A logarithm is defined only when its argument is positive. Therefore:
\[
k(x-h)>0
\]
This inequality gives the domain.
4. Domain of logarithmic functions
If \(k>0\), then:
\[
x-h>0
\]
So:
\[
x>h
\]
If \(k<0\), the inequality reverses:
\[
x
The boundary \(x=h\) is not included because the logarithm argument would be zero.
5. Range of logarithmic functions
A non-constant logarithmic function can output every real number. Therefore, when \(A\neq0\):
\[
\text{Range}=\mathbb{R}
\]
In interval notation:
\[
(-\infty,\infty)
\]
If \(A=0\), the function is constant on its valid domain:
\[
f(x)=v
\]
Then the range is:
\[
\{v\}
\]
6. Asymptotes
Exponential functions have a horizontal asymptote:
\[
y=v
\]
Logarithmic functions have a vertical asymptote:
\[
x=h
\]
The logarithmic asymptote is the boundary where the logarithm argument becomes zero.
7. Worked example
Find the domain and range of:
\[
f(x)=\ln(x-3)+2
\]
The logarithm argument is:
\[
x-3
\]
A logarithm requires a positive argument:
\[
x-3>0
\]
Solve the inequality:
\[
x>3
\]
Therefore, the domain is:
\[
(3,\infty)
\]
Since this is a non-constant logarithmic function, its range is all real numbers:
\[
(-\infty,\infty)
\]
So the final answer is:
\[
\boxed{\text{Domain: }x>3,\quad \text{Range: }\mathbb{R}}
\]
8. Number line interpretation
For \(f(x)=\ln(x-3)+2\), the boundary is \(x=3\).
The point \(3\) is not included because \(\ln(0)\) is undefined.
On a number line, this is shown with an open circle at \(3\) and shading to the right.
9. Summary table
| Function type |
Domain |
Range |
Asymptote |
| \(f(x)=A\cdot b^{k(x-h)}+v\), \(A>0\) |
\(\mathbb{R}\) |
\((v,\infty)\) |
\(y=v\) |
| \(f(x)=A\cdot b^{k(x-h)}+v\), \(A<0\) |
\(\mathbb{R}\) |
\((-\infty,v)\) |
\(y=v\) |
| \(f(x)=A\log_b(k(x-h))+v\), \(k>0\) |
\((h,\infty)\) |
\(\mathbb{R}\), if \(A\neq0\) |
\(x=h\) |
| \(f(x)=A\log_b(k(x-h))+v\), \(k<0\) |
\((-\infty,h)\) |
\(\mathbb{R}\), if \(A\neq0\) |
\(x=h\) |
| Constant special case |
Depends on function type |
\(\{v\}\) or another single value |
May not apply |
10. Common mistakes
- Do not include the boundary \(x=h\) in a logarithmic domain.
- Remember that exponential functions have no input restriction.
- Remember that a non-constant logarithmic function has range \(\mathbb{R}\).
- Do not confuse the vertical shift \(v\) with a domain boundary.
- Always check the logarithm argument, not only the outside expression.
