Totally lost here!
Evaluate the exponential equation for three positive values of x, three negative values of x, and at x=0. Transform the second expression into the equivalent logarithmic equation; and 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.
y=2^x-2,x=2^y-2
To evaluate the exponential equation y = 2^x - 2, we need to substitute different values of x and calculate the corresponding values of y. Let's start by plugging in three positive values of x:
1. For x = 1:
y = 2^1 - 2 = 2 - 2 = 0
2. For x = 2:
y = 2^2 - 2 = 4 - 2 = 2
3. For x = 3:
y = 2^3 - 2 = 8 - 2 = 6
Now, let's substitute three negative values of x:
1. For x = -1:
y = 2^-1 - 2 = 1/2 - 2 = -3/2
2. For x = -2:
y = 2^-2 - 2 = 1/4 - 2 = -7/4
3. For x = -3:
y = 2^-3 - 2 = 1/8 - 2 = -15/8
Finally, we evaluate the equation at x = 0:
y = 2^0 - 2 = 1 - 2 = -1
Next, let's transform the second expression, x = 2^y - 2, into an equivalent logarithmic equation. To do this, we swap the roles of x and y:
y = 2^x - 2 is equivalent to x = log(base 2)(y + 2)
Now, we can evaluate the logarithmic equation for three values of x that are greater than 1:
1. For x = 2:
y = log(base 2)(2 + 2) = log(base 2)(4) = 2
2. For x = 3:
y = log(base 2)(3 + 2) = log(base 2)(5)
3. For x = 4:
y = log(base 2)(4 + 2) = log(base 2)(6)
Similarly, let's evaluate the logarithmic equation for three values of x that are between 0 and 1:
1. For x = 1/2:
y = log(base 2)((1/2) + 2) = log(base 2)(2.5)
2. For x = 1/4:
y = log(base 2)((1/4) + 2) = log(base 2)(2.25)
3. For x = 1/8:
y = log(base 2)((1/8) + 2) = log(base 2)(2.125)
Finally, let's evaluate the equation at x = 1:
y = log(base 2)(1 + 2) = log(base 2)(3)
These are the calculations based on the given equations.
To evaluate the exponential equation, y = 2^(x-2), for the given values of x, you need to substitute those values into the equation and calculate the corresponding values of y.
Positive values of x:
Let's take three positive values of x, for example, x = 1, x = 2, and x = 3.
For x = 1:
y = 2^(1-2)
y = 2^(-1)
y = 1/2
For x = 2:
y = 2^(2-2)
y = 2^0
y = 1
For x = 3:
y = 2^(3-2)
y = 2^1
y = 2
So, the corresponding values of y are:
For x = 1, y = 1/2
For x = 2, y = 1
For x = 3, y = 2
Negative values of x:
Similarly, let's take three negative values of x, for example, x = -1, x = -2, and x = -3.
For x = -1:
y = 2^(-1-2)
y = 2^(-3)
y = 1/8
For x = -2:
y = 2^(-2-2)
y = 2^(-4)
y = 1/16
For x = -3:
y = 2^(-3-2)
y = 2^(-5)
y = 1/32
So, the corresponding values of y for negative x values are:
For x = -1, y = 1/8
For x = -2, y = 1/16
For x = -3, y = 1/32
x = 0:
For x = 0:
y = 2^(0-2)
y = 2^(-2)
y = 1/4
The corresponding value of y for x = 0 is y = 1/4.
Now, let's transform the second expression, y = 2^(x-2), into an equivalent logarithmic equation.
y = 2^(x-2)
Logarithmic equation: log2(y) = (x-2)
To evaluate the logarithmic equation, log2(y) = (x-2), for the given values of x, you need to substitute those values into the equation and calculate the corresponding values of log2(y).
Values of x greater than 1:
Let's take three values of x that are greater than 1, for example, x = 2, x = 3, and x = 4.
For x = 2:
log2(y) = (2-2)
log2(y) = 0
y = 2^0
y = 1
For x = 3:
log2(y) = (3-2)
log2(y) = 1
y = 2^1
y = 2
For x = 4:
log2(y) = (4-2)
log2(y) = 2
y = 2^2
y = 4
So, the corresponding values of y for x values greater than 1 are:
For x = 2, y = 1
For x = 3, y = 2
For x = 4, y = 4
Values of x between 0 and 1:
Similarly, let's take three values of x that are between 0 and 1, for example, x = 0.25, x = 0.5, and x = 0.75.
For x = 0.25:
log2(y) = (0.25-2)
log2(y) = -1.75
y = 2^(-1.75)
For x = 0.5:
log2(y) = (0.5-2)
log2(y) = -1.5
y = 2^(-1.5)
For x = 0.75:
log2(y) = (0.75-2)
log2(y) = -1.25
y = 2^(-1.25)
Finally, let's evaluate the logarithmic equation for x = 1:
For x = 1:
log2(y) = (1-2)
log2(y) = -1
y = 2^(-1)
So, the corresponding values of y for x = 1 is y = 1/2.
I hope this explanation helps you understand how to evaluate the given exponential and logarithmic equations and solve for the given values of x.