Any math pro can solve this???

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Please solve and tell working!!! :C

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 of \(n \).
 
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PlanetMaster

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This should be straightforward.

We have y=4 when x=3, so
\(4=a\cdot3^{n}-23 \)
\(a=\frac{27}{3^{n}} \)


We also have y=220 when x=9, so
\(220=a\cdot9^{n}-23 \)
\(a=\frac{243}{9^{n}} \)


Equating a from both equations, we get
\(\frac{27}{3^{n}}=\frac{243}{9^{n}} \)
\(\frac{9^{n}}{3^{n}}=\frac{243}{27} \)
\((\frac{9}{3})^{n}=9 \)
\(3^{n}=9 \)


We can use log rule here (and log power rule) to find n
\(\log 3^{n}=\log 9 \)
\(n\log 3=\log 9 \)
\(n=\frac{\log 9}{\log 3} \)
\(n=2 \)


Substituting n in either of the a equation above gives us
\(a=\frac{27}{3^{n}} \)
\(a=\frac{27}{3^2} \)
\(a=\frac{27}{9} \)
\(a=3 \)


Hope this helps!
 
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Doesn't this method just ignore the -23?
You can substitute known y and x values and rearrange to get the two equations:
27=a*3^n and 243=a*9^n ( simplifies to 243=a*3^2n as 9=3^2)
Divide second equation by first to get 9=3^n and so n=2.
Substitute n back into 27=a*3^2 to find that a=3.
 

PlanetMaster

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Doesn't this method just ignore the -23?
You can substitute known y and x values and rearrange to get the two equations:
27=a*3^n and 243=a*9^n ( simplifies to 243=a*3^2n as 9=3^2)
Divide second equation by first to get 9=3^n and so n=2.
Substitute n back into 27=a*3^2 to find that a=3.
Ah damn, I missed that! Good catch there..
I was a bit skeptical of a decimal answer at first as these type of questions in GCSE/O-Levels usually have whole numbers as answers but missed that I guess.

Anyway, I've made the corrections in the my answer above as well.
Thanks! :)
 
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