Exercise 44 Page 484

We want to find an exponential function that passes through (2,3) and (5, 19). To do this, we will create a system of equations by substituting ( x, y) = ( 2, 3) into the general form of exponential function, y=ab^x, and then substituting again with ( x, y) = ( 5, 19). 3=a * b^2 & (I) 19=a * b^5 & (II) We want to solve the above system of equations. When solving a system of equations using substitution, there are three steps.

1. Isolate a variable in one of the equations.
2. Substitute the expression for that variable into the other equation and solve.
3. Substitute this solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate a in the first equation.

3=a * b^2 & (I) 19=a * b^5 & (II)

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(I): Solve for a

DivEqn

(I): .LHS /b^2.=.RHS /b^2.

3b^2=a 19=a * b^5

RearrangeEqn

(I): Rearrange equation

a= 3b^2 19=a * b^5

Now that we have isolated a we can solve the system by substitution.

a= 3b^2 19=a * b^5

Substitute

(II):a= 3/b^2

a= 3b^2 19= 3b^2 * b^5

SplitIntoFactors

(II): Split into factors

a= 3b^2 19= 3b^2 * b^2 * b^3

FracMultDenomToNumber

(II): a/b^2* b^2 = a

a= 3b^2 19=3 * b^3

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(II): Solve for b

DivEqn

(II): .LHS /3.=.RHS /3.

a= 3b^2 127=b^3

RadicalEqn

(II): sqrt(LHS)=sqrt(RHS)

a= 3b^2 sqrt(127)=sqrt(b^3)

CalcRoot

(II): Calculate root

a= 3b^2 13=b

RearrangeEqn

(II): Rearrange equation

a= 3b^2 b= 13

Great! Now, to find the value of a we need to substitute b= 13 into either one of the equations in the given system. Let's use the first equation.

a= 3b^2 b= 13

Substitute

(I):b= 1/3

a= 3( 1/3)^2 b= 13

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Evaluate

CalcPow

(I): Calculate power

a= 31/9 b= 13

DivByFracD

(I): a/b/c= a * c/b

a= 271 b= 13

DivByOne

(I): a/1=a

a=27 b= 13

The solution to this system of equations is a=27 and b=13. With this information, we can finally write an equation of the exponential function that passes through the given points. y=ab^x ⇔ y=27 (1/3)^x