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

1 Answer

9 votes

Answer:


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).

User Evalsyrelec
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