# Any math pro can solve this???

#### Ahmed Iqbal123

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

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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#### Ahmed Iqbal123

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#### PG MathsXPhysics

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

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!

#### Ahmed Iqbal123

i knew you missed it but i still understood