Question 976027
{{{y = ax^n + bx}}}
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When {{{x=1}}},
{{{4 = a + b}}}
1. {{{a=4-b}}}
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When {{{x=3}}},
2.{{{6=a*3^n+3b}}}
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When {{{x=9}}},
{{{15=a*9^n+9b}}}
3.{{{15=a*3^(2n)+9b}}}
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From eq. 2,
{{{a*3^n=3b-6}}}
{{{a^2*3^(2n)=9(b-2)^2}}}
From eq. 3,
{{{a*3^(2n)+9b=15}}}
So,
{{{a^2*3^(2n)+9ab=15a}}}
Comparing,
{{{9(b-2)^2=15a-9ab}}}
{{{9(b-2)^2=3a(5-3b)}}}
So then substituting from eq. 1,
{{{9(b-2)^2=3(4-b)(5-3b)}}}
{{{9(b^2-4b+4)=3(20-12b-5b+3b^2)}}}
{{{9b^2-36b+36=60-51b+9b^2}}}
{{{15b-24=0}}}
{{{15b=24}}}
{{{b=24/15}}}
{{{highlight(b=8/5)}}}
Then,
{{{a=4-8/5}}}
{{{a=20/5-8/5}}}
{{{highlight(a=12/5)}}}
And finally,
{{{6=(12/5)*3^n+3(8/5)}}}
{{{30=12*3^n+24}}}
{{{6=12*3^n}}}
{{{3^n=1/2}}}
{{{n=log(3,(1/2))}}}
{{{n=log((1/2))/log((3))}}}
{{{highlight(n=-log((2))/log((3)))}}}