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How would you go about converting ${1\over y} = x^2 + a^{b-x} $ into linear form. I know how you would normally go about solving this type of problem but I fail to make any progress on this one.

I have to make it in the form $Y = mX + c$

where, $Y$ and $X$ are in terms of $x$ and $y$ (the variables) without a and b (the constants); m and c are in terms of a and b (the constants) without x and y (the variables).

I'm having trouble separating the $a$ and $x$ in the $a^{b-x} $ term. Thank you, all help is appreciated.

note: I haven't posted any of my working because I wasn't really able to get anywhere.

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1 Answer

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Ok I was able to solve it, in case anyone was wondering...

${1\over y} = x^2 + a^{b-x}$

${1\over y}- x^2= a^{b-x}$

$\log({{1\over y}- x^2})= \log({a^{b-x}})$

$\log({{1\over y}- x^2})= (b-x)\log({a})$

$\log({{1\over y}- x^2})= -\log({a})\times x + \log{a^b}$

$Y = mX + c$

where $Y = \log({{1\over y}- x^2})$ , $m=-\log({a})$ , $X = x$ and $c=\log{a^b}$

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