Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
2. Section 2.2
Continue to next subchapter

Exercise 110 Page 93

Practice makes perfect
a The general form of an exponential function is written in the following format.
y=ab^x To find the equation, we need to determine a and b. From the exercise, we know that the function passes through two points. This means we can substitute both of these points in the function creating two equations. |c|c| [-0.8em] Point & y=ab^x [0.4em] [-0.8em] ( 3, 0.05) & 0.05=ab^3 [0.4em] [-0.8em] ( 5, 0.0125) & 0.0125=ab^5 [0.4em] If we combine these, we get a system of equations which we can be solved by using the Substitution Method.
0.05=ab^3 & (I) 0.0125=ab^5 & (II)
â–Ľ
(I): Solve for a
0.05b^3=a 0.0125=ab^5
a= 0.05b^3 0.0125=ab^5
a= 0.05b^3 0.0125= 0.05b^3* b^5
â–Ľ
(II): Solve for b
a= 0.05b^3 0.0125= 0.05b^5b^3
a= 0.05b^3 0.0125=0.05b^2
a= 0.05b^3 0.25=b^2
a= 0.05b^3 b^2=0.25
a= 0.05b^3 b=± 0.5

(II): b > 0

a= 0.05b^3 b=0.5
Notice that b must be non-negative since we cannot have a negative base in an exponential function. To find a, we substitute the value of b back into the first equation and evaluate.
a= 0.05b^3 b=0.5
a= 0.05 0.5^3 b=0.5
â–Ľ
(I): Evaluate right-hand side
a= 0.050.125 b=0.5
a=0.4 b=0.5
Now we can complete the equation. y=0.4(0.5)^x
b Like in Part A, we have to substitute the known points into the general form of an exponential function.
|c|c| [-0.8em] Point & y=ab^x [0.4em] [-0.8em] ( 1, 16) & 16=ab^1 [0.4em] [-0.8em] ( 4, 128) & 128=ab^4 [0.4em] If we combine these, we get a system of equations which can be solved by using the Substitution Method.
16=ab^1 & (I) 128=ab^4 & (II)
â–Ľ
(I): Solve for a
16=ab 128=ab^4
ab=16 128=ab^4
a= 16b 128=ab^4
a= 16b 128=( 16b)b^4
â–Ľ
(II): Solve for b
a= 16b 128= 16b^4b
a= 16b 128=16b^3
a= 16b 8=b^3
a= 16b b^3=8
a= 16b b=2
To find a, we substitute the value of b back into the first equation and evaluate.
a= 16b b=2
a= 16 2 b=2
a=8 b=2
Now we can complete the equation. y=8(2)^x