# Suppose that  f  is an exponential function with a percentage growth rate of  2% , and with f(0)=147. Find a formula for  f .a) f(t)=0.02t+147b) f(t)=1.02(1.47)tc) f(t)=147(1.02)td) f(t)=147(2)te) f(t)=147(1.20)t

Suppose that  f  is an exponential function with a percentage growth rate of  2% , and with $f\left(0\right)=147$. Find a formula for  f .
a) $f\left(t\right)=0.02t+147$
b) $f\left(t\right)=1.02\left(1.47\right)t$
c) $f\left(t\right)=147\left(1.02\right)t$
d) $f\left(t\right)=147\left(2\right)t$
e) $f\left(t\right)=147\left(1.20\right)t$

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Alannej

We have to find formula for f if growth rate is 2% and $f\left(0\right)=147$
Definition of exponential function:
An exponential growth or decay function is the function which grows or shri
at some condition of growth percentage,
so equation of exponential function can be expressed as:
$f\left(x\right)=a\left(1+r{\right)}^{x}orf\left(x\right)=a{b}^{x}$ (took $b=1+r$)
Here,
a is the initial value
r is the growth or decay rate written in decimal
b is the factor of growth or decay
According to question,
$r=2\mathrm{%}$
=$2/100$)
$=0.02$
so,
$b=1+r$
$=1+0.02$
$=1.02$
Step 2
Taking variable as t,
$f\left(t\right)=a{b}^{t}$
Therefore,
$f\left(t\right)=abt=a\left(1.02{\right)}^{t}$
Given, $f\left(0\right)=147$
putting $t=0$, we get
$f\left(t\right)=a\left(1.02{\right)}^{t}$
$f\left(0\right)=a\left(1.02{\right)}^{0}$
$147=a\cdot 1$
$a=147$
${\text{(since, (any finite number)}}^{0}=1\right)$
After putting value of a we can write the formula for f,
$f\left(t\right)=147\left(1.02{\right)}^{t}$
Hence, option c) is correct.