Question

Write an exponential function in the form y=ab^x that goes through points (0, 20) and (7, 2560).

Answer 1

`y = 20* 2^(x)`.

Explanation:

Let
`y = (a)\, (b)^(x)` denote this exponential function for some constant
`a` and
`b` (
`b > 0`) that need to be found.

Assume that the graph of the function
`y = (a)\, (b)^(x)` goes through the point
`(x_(0),\, y_(0))`. Substituting in
`x = x_(0)` and
`y = y_(0)` should satisfy the equation of this function:

`y_(0) = (a)\, (b)^{x_(0)}`.

For example,
`y = (a)\, (b)^(x)` goes through the point
`(0,\, 20)`. Substituting in
`x = 0` and
`y = 20` should then satisfy the equation of this function:

`20 = (a)\, (b)^(0)`.

Similarly, since the graph of this function goes through
`(7,\, 2560)` where
`x = 7` and
`y = 2560`:

`2560 = (a)\, (b)^(7)`.

In general, the constant
`a` in such systems could be eliminating by dividing one equation by the other. For example, dividing
`2560 = (a)\, (b)^(7)` by
`20 = (a)\, (b)^(0)` gives:

`\displaystyle (2560)/(20) = ((a)\, (b)^(7))/((a)\, (b)^(0))`.

Simplify to obtain:

`\displaystyle 128 = b^(7 - 0)`.

`\displaystyle 128 = b^(7)`.

`b^(7) = 128`.

Since the exponent of
`b` is an odd number, the sign of
`b` should also be positive- same as the sign of
`128`. Extra steps would be required if the exponent of
`b` is even.

Raise both sides to the
`(1/7)`th power to find the (unique) value of
`b`:

`b = 128^(1/7) = 2`.

Substitute
`b = 2` back into either equation (for example,
`2560 = (a)\, (b)^(7)`) to find the value of
`a`:

`2560 = (a)\, (2)^(7)`.

`\begin{aligned}a &= (2560)/(2^(7)) \\ &= (2560)/(128) \\ &= 20\end{aligned}`.

Thus, this exponential function would be
`y = 20* 2^(x)`.