Question

Hi I need the answer to the question quickly if possible please

Answer 1

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 $g(x) = log_{4}(x-1)$

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.

$f(2) = log_{4}((2) - 1)$

Simplify the result

Subtract 1 from 2

$f(2) = log_{4}(1)$

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.

$f(2) = log_{4}((5) - 1)$

$f(2) = log_{4}(4)$

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.

$f(2) = log_{4}((3) - 1)$

$f(2) = log_{4}(2)$

Logarithm base 4 of 2 is $\frac{1}{2}$.

Rewrite as an equation.

$log_{4}((2) = x$

Rewrite $log_{4}((2) = x$ in exponential form using definition of a logarithm. If x and b are positive real numbers and b does not equal 1, then $log_{b}((x) = y$ is equivalent to $b^{y} = x.$

$4^{x} = 2$

Create expressions in the equation that all have equal bases.

$(2^{2})^{x} = 2^{1}$

Rewrite $(2^{2})^{x} as 2^{2x}$

$2^{2x} = 2^{1}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

2x = 1

solve for x

$x = \frac{1}{2}$

The variable x is equal to $\frac{1}{2}$

$f(3) = \frac{1}{2}$

The final answer is $\frac{1}{2}$

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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