Linear and Exponential

[suicoted]

Given the values:

x | 2 | 5 | 8
-----------------
y | 64 | ? | 25

Find y(5) if y is an exponential function of x.
So do I just go 8-2/25-64?

 

[stapel]

To evaluate f(5), you might want to find the formula for f(x). I'm not sure why you think the slope between two other points on the curve would give you the value of y at x = 5. (Would the slope give you the value of y for any x between 2 and 8? If so, why? If not, then why would this work for x = 5?)

You are told that y is exponential, so you know that is it of the form y = ab^x, where b is the base and a is some multiplier. You are given two data points, (2, 64) and (8, 25). Plug them in, and solve the system:

. . . . .ab² = 64
. . . . .ab⁸ = 25

Then:

. . . . .ab⁸ = ab²b⁶ = 25 = 64b⁶
. . . . .25/64 = b⁶
. . . . .5/8 = b³
. . . . .cbrt(5/8) = b = cbrt(5)/2

. . . . .a(cbrt(5)/2)² = 64
. . . . .a(cbrt(25)/4) = 64
. . . . .a = 256/cbrt(25)

...and so forth.

Eliz.

 

[suicoted]

To evaluate f(5), you might want to find the formula for f(x). I'm not sure why you think the slope between two other points on the curve would give you the value of y at x = 5. (Would the slope give you the value of y for any x between 2 and 8? If so, why? If not, then why would this work for x = 5?)

You are told that y is exponential, so you know that is it of the form y = ab^x, where b is the base and a is some multiplier. You are given two data points, (2, 64) and (8, 25). Plug them in, and solve the system:

. . . . .ab² = 64
. . . . .ab⁸ = 25

Then:

. . . . .ab⁸ = ab²b⁶ = 25 = 64b⁶
. . . . .25/64 = b⁶
. . . . .5/8 = b³
. . . . .cbrt(5/8) = b = cbrt(5)/2

. . . . .a(cbrt(5)/2)² = 64
. . . . .a(cbrt(25)/4) = 64
. . . . .a = 256/cbrt(25)

...and so forth.

Eliz.

What's cbrt? I don't get the last three steps.

 

[stapel]

"Cube root of x" is customarily expressed as "cbrt(x)", similar to how "square root of x" is formatted as "sqrt(x)".

The last three steps show how to plug the known value for b into the first of the two original equations, to start solving for a.

Eliz.

 

[suicoted]

"Cube root of x" is customarily expressed as "cbrt(x)", similar to how "square root of x" is formatted as "sqrt(x)".

The last three steps show how to plug the known value for b into the first of the two original equations, to start solving for a.

Eliz.

So to solve for 5, I go

(256/cbrt(25))(cbrt(5)/2)^6 =(approx) 34.199?
If so, thanks so much.

I'm curious, how did u come up with:
"ab^2b^6" in the first step? How did u know what two numbers to break up for b? You could have chosen 4 and 4, or 7 and 1, or 5 and 3.

 

[stapel]

how did u come up with "ab^2b^6" in the first step?

When you have a non-linear system, substitution is generally the way to go to solve it. I had ab² and ab⁸, and the above seemed an obvious way to relate the two.

Eliz.