Evaluate the logarithmic equation for three values of x that are greater than 1, three values of x that are between 0 and 1, and at x=1.
show your work.
use the resulting ordered pairs to plot the graph.
State the equation of the line asymptotic to the graph (if any)
y= -log5 ^x
please help I am very lost on this problem. Thanks
Y = -Log(5^x)
Y = -Log(5^0.5) = -Log(2.236)=- 0.3495.
Y = -Log(5^1) = -Log(5) = -0.6990.
Y = -Log(5^2) = -Log(25) = - 1.3979.
(0.5 , -0.35) , (1.0 ,- 0.70) , (2.0 , -1.40).
I did one of each; you can add the remainihg points.
The asymptote is 0 for x > 1 and -1 for x in [0,1].
To evaluate the logarithmic equation for different values of x, we substitute the values of x into the equation and compute the corresponding values of y.
The given equation is:
y = -log5(x)
1. Values of x > 1:
Let's choose three values for x that are greater than 1, such as 2, 3, and 4.
For x = 2:
y = -log5(2)
To evaluate this logarithm, we can use the change of base formula to convert it into base 10 logarithm.
y = -log10(2) / log10(5)
Using a calculator, we find that y ≈ -0.4307.
For x = 3:
y = -log5(3)
Similarly, using the change of base formula, we get:
y ≈ -0.6826.
For x = 4:
y = -log5(4)
Applying the change of base formula, we obtain:
y ≈ -0.8614.
So, for x > 1, the values of y are approximately -0.4307, -0.6826, and -0.8614. These three pairs of coordinates are (2, -0.4307), (3, -0.6826), and (4, -0.8614).
2. Values of x between 0 and 1:
Let's choose three values for x that are between 0 and 1, such as 0.2, 0.3, and 0.5.
For x = 0.2:
y = -log5(0.2)
Following the change of base formula:
y ≈ -0.4307.
For x = 0.3:
y = -log5(0.3)
Again, evaluating with the change of base formula:
y ≈ -0.6826.
For x = 0.5:
y = -log5(0.5)
Using the change of base formula, we find:
y ≈ -1.
So, for x between 0 and 1, the values of y are approximately -0.4307, -0.6826, and -1. These three pairs of coordinates are (0.2, -0.4307), (0.3, -0.6826), and (0.5, -1).
3. Value of x = 1:
For x = 1:
y = -log5(1)
The logarithm of 1 to any base is always 0. So, y = 0.
For x = 1, the value of y is 0.
To plot the graph, we place the values obtained in the xy-plane:
(2, -0.4307), (3, -0.6826), (4, -0.8614), (0.2, -0.4307), (0.3, -0.6826), (0.5, -1), and (1, 0).
Now, regarding the line asymptotic to the graph, we observe that as x approaches positive infinity, the value of y approaches negative infinity. Therefore, there is a vertical asymptote at x = +∞.