Find an approimate equation...

• March 7th 2008, 01:08 AM
mt_lapin
Find an approimate equation...
Find an approximate equation $y=ab^{x}$ of the exponential curve that contains the given pair of points. Round the value of b to two decimal places.

(0, 256) and (7, 23)

my work:

$256b^7=23$

$256^7=\frac{23}{256}$

$b=\sqrt[7]{\frac{23}{256}}$ which is $b=4.7$

Thus, $y=256(4.7)^{x}$

• March 7th 2008, 01:19 AM
mr fantastic
Quote:

Originally Posted by mt_lapin
Find an approximate equation $y=ab^{x}$ of the exponential curve that contains the given pair of points. Round the value of b to two decimal places.

(0, 256) and (7, 23)

my work:

$256b^7=23$

$256^7=\frac{23}{256}$

$b=\sqrt[7]{\frac{23}{256}}$ Mr F says: Correct.

which is $b=4.7$ Mr F says: Time to get a new calculator. The seventh root of 23/256 cannot possibly be greater than 1! I get 0.71, correct to two decimal places.

Thus, $y=256(4.7)^{x}$

I did $\left(\frac{23}{256}\right)^7$ when I should have done $\left(\frac{23}{256}\right)^{1/7}$ ... I made a mistake.