Question 818780
Given that y=ax^n - 23 and that y=4 when x=3 and y=220 when x=9 find the value of a and n.

y=ax^n - 23 and that y=4 when x=3 and y=220 when x=9 find the value of a and n.

4 = a(3)^n  
and 
220 = a(9)^n

Let's solve each equation for a
4 = a(3)^n
divide each side by (3)^n an we have
{{{a=4/((3)^n)}}}
Similarly 220 = a(9)^n can be written
{{{a=220/((9)^n)}}}
Setting the right hand sides to be equal,
{{{4/((3)^n)}}}={{{220/((9)^n)}}}
Taking the cross products we have
{{{4*((9)^n)}}}={{{220*((3)^n)}}}
Divide each side by 4
{{{((9)^n)}}}={{{(220/4)*((3)^n)}}}or{{{((9)^n)}}}={{{55*((3)^n)}}}
Divide each side by {{{((3)^n)}}}
= {{{((9)^n)/((3)^n)}}}={{{55}}}or {{{((3^2)^n)/((3)^n)}}}={{{55}}}
= {{{((3^n)^2)/((3)^n)}}}={{{55}}}So,{{{(3)^n}}}={{{55}}}
Take the log of each side
{{{log((3)^n)}}}={{{log(55)}}}so, {{{n*log((3))}}}={{{log(55)}}}
Divide each side by log(3)
{{{n}}}={{{(log(55))/(log(3))}}}=3.647632
Now substitute in {{{a=4/((3)^n)}}},{{{a=4/((3)^3.647632)}}},{{{a=4/(55)}}}
a=0.072727
Let's check the second equation, 220 = a(9)^n
220 ?= (0.072727)(9^3.647632)
220 ?= (0.072727)(3025.001)
     = 219.9992   So considering rounding error we seem to be correct.