The log function can be graphed using the vertical asymptote at x = 1 and the points (2,0), (5,1) & (3,0.5).
Given graph is ![g(x) = log_{4}(x-1)](https://tex.z-dn.net/?f=g%28x%29%20%3D%20log_%7B4%7D%28x-1%29)
We have to find the asymptotes.
Set the argument of the logarithm equal to zero.
x - 1 = 0
Now add 1 to both the sides of the equation.
x - 1 + 1 = 0 + 1
= x = 1
The vertical asymptote occurs at x = 1
So, vertical asymptote: x = 1
Now, find the point at x = 2
Replace the variable x with 2 in the expression.
![f(2) = log_{4}((2) - 1)](https://tex.z-dn.net/?f=f%282%29%20%3D%20log_%7B4%7D%28%282%29%20-%201%29)
Simplify the result
Subtract 1 from 2
![f(2) = log_{4}(1)](https://tex.z-dn.net/?f=f%282%29%20%3D%20log_%7B4%7D%281%29)
Logarithm base 4 of 1 is 0
f(2) = 0
The final answer is 0.
y = 0
Now find the point at x = 5
Replace the variable x with 5 in the expression.
![f(2) = log_{4}((5) - 1)](https://tex.z-dn.net/?f=f%282%29%20%3D%20log_%7B4%7D%28%285%29%20-%201%29)
![f(2) = log_{4}(4)](https://tex.z-dn.net/?f=f%282%29%20%3D%20log_%7B4%7D%284%29)
Logarithm base 4 of 4 is 1
so, f(5) = 1
y = 1
Now find the point at x = 3
Replace the variable x with 3 in the expression.
![f(2) = log_{4}((3) - 1)](https://tex.z-dn.net/?f=f%282%29%20%3D%20log_%7B4%7D%28%283%29%20-%201%29)
![f(2) = log_{4}(2)](https://tex.z-dn.net/?f=f%282%29%20%3D%20log_%7B4%7D%282%29)
Logarithm base 4 of 2 is
.
Rewrite as an equation.
![log_{4}((2) = x](https://tex.z-dn.net/?f=log_%7B4%7D%28%282%29%20%3D%20x)
Rewrite
in exponential form using definition of a logarithm. If x and b are positive real numbers and b does not equal 1, then
is equivalent to ![b^{y} = x.](https://tex.z-dn.net/?f=b%5E%7By%7D%20%3D%20x.)
![4^{x} = 2](https://tex.z-dn.net/?f=4%5E%7Bx%7D%20%3D%202)
Create expressions in the equation that all have equal bases.
![(2^{2})^{x} = 2^{1}](https://tex.z-dn.net/?f=%282%5E%7B2%7D%29%5E%7Bx%7D%20%3D%202%5E%7B1%7D)
Rewrite ![(2^{2})^{x} as 2^{2x}](https://tex.z-dn.net/?f=%282%5E%7B2%7D%29%5E%7Bx%7D%20as%202%5E%7B2x%7D)
![2^{2x} = 2^{1}](https://tex.z-dn.net/?f=2%5E%7B2x%7D%20%20%3D%202%5E%7B1%7D)
Since the bases are the same, then two expressions are only equal if the exponents are also equal.
2x = 1
solve for x
![x = \frac{1}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
The variable x is equal to ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
![f(3) = \frac{1}{2}](https://tex.z-dn.net/?f=f%283%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
The final answer is ![\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D)
So, y = 0.5
The log function can be graphed using the vertical asymptote at x = 1 and the points (2,0), (5,1) & (3,0.5).
x y
2 0
3 0.5
5 1
Hence the answer is the log function can be graphed using the vertical asymptote at x = 1 and the points (2,0), (5,1) & (3,0.5).
To learn more about graphs, click here brainly.com/question/19040584
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