If a function contains the points (1, 3), (2, 9), and (3, 27), is it most likely exponential or linear?

First, recognize that with every increase in x, y is multiplied by 3 from the previous value, rather than adding the same value. This means that it's probably exponential, not linear.

Solve for y:

3^{(2x + 4)} = 9^y

First, find the common base:

3^{(2x + 4)} = (3²)^y

Use exponential rules:

3^{(2x + 4)} = 3^2y

Now we've got the same base on both sides, so throw them straight into the garbage:

2x + 4 = 2y

Solve for y:

2y = 2x + 4

y = x + 2

Solve for x:

7^{(7x – 3)} = 49^-2x

Notice that both sides have a common factor of 7:

7^{(7x – 3)} = (7²)^-2x

Multiply those exponents on the right side:

7^{(7x – 3)} = 7^-4x

Hey, look at that: matching bases. Now we can drop both bases:

7x – 3 = -4x

Solve for x:

11x = 3

Solve for x without using a calculator:

We've got a negative exponent on the left, which means we can bring the denominator up into the numerator by swapping the sign on that exponent:

(27^{7x + 1})(81^4x) = 9

Find the common base:

[(3³)^{7x + 1}][(3⁴)^4x] = 9

Yeesh, that looks gnarly. Let's multiply the exponents:

(3^{21x + 3})(3^16x) = 9

Now we can add the exponents on like bases:

3^{37x + 3} = 9

Turn that 9 into 3² to give us matching bases on both sides:

3^{37x + 3} = 3²

Finally, we can drop our bases:

37x + 3 = 2

Finish up by solving for x, algebra-style:

37x = -1

Find an exponential function f (x) = Cb^x that contains the following set of points: {(1, 4), (2, 8), (5, 64)}.

First, note that between x = 1 and x = 2, y doubles. Because each increase in x means multiplying the answer by another b, we know that b = 2.

f (x) = C(2)^x

Now plug in any of those three points to find C. We'll go with (1, 4), which means f (1) = 4.

f (x) = C(2)^x

4 = C(2)¹

4 = 2C

Solve for C:

C = 2

Sweet, now we've got our coefficient and our base. That means our function must be f (x) = 2(2)^x. We still need to make sure it matches the other two points, though. First we try (2, 8):

f (x) = 2(2)^x

f (2) = 2(2)² = 2(4) = 8

Yep, that worked. Now for (5, 64):

f (x) = 2(2)^x

f (5) = 2(2)⁵ = 2(32) = 64

Perfect. We're done.