1

Reduce the following equation to a linear relationship:

$y-2000=ab^{-x}$

The way I've seen to do this so far is to apply log to each side, and use that to get the equation into the form $Y=mX+c$. I'm not entirely sure how I could do that here; I've tried the following;

Rearrange the formula:

$y=ab^{-x}+2000$

Log each side:

$log(y)=log(ab^{-x} +2000)$

As far as I know, the right hand side can't be simplified any further, and so I'm at a bit of a loss as to how to reduce this to a linear relationship.

I have been trying to think of where to go with the problem but I'm at a dead end. Can anyone offer any help?

OliverRadini
  • 153

1 Answers1

1

If you subtract both sides by 2000 and then multiply through by $b^{x}$ you get $(y-2000)b^{x} = a$, and then applying $\log$ to both sides, you get $\log(y-2000) + \log(b^{x}) = \log(a)$. Assuming that $a$ and $b$ are constants, that can be re-rerwitten as $\log(b) x - \log(a) = - \log(y-2000)$.

That looks like the best you can do to me.

AshSeifert
  • 328