Question

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

Answer 1

Answer:

$y = 20\times 2^{x}$.

Step-by-step 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:

$\displaystyle \frac{2560}{20} = \frac{(a)\, (b)^{7}}{(a)\, (b)^{0}}$.

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 $a$:

$2560 = (a)\, (2)^{7}$.

$\begin{aligned}a &= \frac{2560}{2^{7}} \\ &= \frac{2560}{128} \\ &= 20\end{aligned}$.

Thus, this exponential function would be $y = 20\times 2^{x}$.