Question

Write an equation for the transformed logarithm shown below, that passes through (-2,0) and (2,-2)

Answer 1

Answer:

$y=\log_{\frac{1}{2}}(x+2).$

Step-by-step explanation:

Find the equation of the transformed logarithm as $y=\log_a (x-b).$

For the point (-2,0):

$0=\log_a (-2-b).$

For the point (2,-2):

$-2=\log_a (2-b).$

Solve the system of equations:

$\left\{\begin{array}{l}0=\log_a (-2-b)\\-2=\log_a (2-b)\end{array}\right.\Rightarrow \left\{\begin{array}{l}b=-2\\-2=\log_a 4\end{array}\right.$

Then

$a^{-2}=a^{\log_a 4} \Rightarrow \dfrac{1}{a^2}=4,\ a=\dfrac{1}{2}.$

The equation of the function is $y=\log_{\frac{1}{2}}(x+2).$

Answer 2

The correct answer is: $f(x)=\frac{-2log(x+3)}{log(5)}$

(In compact form: $f(x) = -2log_5(x+3)$)

Explanation:

Follow these steps...

1. As it is a logarithmic graph, but reflected along x-axis, the general form of the function will be the following:

f(x) = -a*log(x) + k --- (A)

Where a represent the scale of the logarithmic graph, and k represents whether the graph is shifted down or up.

The negative sign indicates that the graph is reflected along the x-axis.

2. The vertical asymptote is at x = -3, the equation (A) will now become the following:

f(x) = -a*log(x + 3) + k --- (B)

The (x+3) part of the above equation shows that the graph is shifted 3 units towards left.

3. Now we need to find k and a. For finding k, plug in the point (-2,0) in equation (B):

0 = -a*log(-2+3) + k

0 = -a*0 + k

k = 0

4. For finding a, plug in the point (2,-2) and the value of k=0 in equation (B):

-2 = -a*log(2+3) + 0

a = 2/log(5)

Now plug in the values of a and k in equation (B), you will get the answer:

f(x) = -(2/log(5))*log(x+3) + 0

Therefore, the correct answer is:

$f(x)=\frac{-2log(x+3)}{log(5)}$

or, in compact form:

$f(x) = -2log_5(x+3)$