An equation with exponentials and would like to rewrite for x.
$f= \frac{-1}{(10^{x-a}+1)(10^{x-b}+1)}$
$\frac{-1}{f}= {(10^{x-a}+1)(10^{x-b}+1)}$
f, a, b are known values and can variate. And plugin some values for it, results in following form, e.g.
$\frac{-1}{f}= {(10^{x-9.35}+1)(10^{x-5.22}+1)}$
$\frac{-10^0}{f}= {(10^{x-9.35}+10^0)(10^{x-5.22}+10^0)}$
$\frac{-10^0}{f}= {(10^{2x-9.35-5.22} + 10^{x-9.35} + 10^{x-5.22} + 10^0)}$
$\frac{-10^0}{f}= {(10^{2x-9.35-5.22} + 10^{x-9.35} + 10^{x-5.22} + 10^0)}$
${1 = -f(10^{2x-9.35-5.22} + 10^{x-9.35} + 10^{x-5.22} + 10^0)}$
${0 = -f*10^{2x-9.35-5.22} -f*10^{x-9.35} -f*10^{x-5.22} -f*10^0 -1}$
${0 = 10^{2x-4.12} +10^{x-9.35} +10^{x-5.22} +10^0 +\frac{1}{f}}$
I tried to look at it in terms of quadratic form. However it doesn't seem right.
I would like some help to rewrite the equation for x using f, a, b and then solve for x?
Start from $$-1/f = (10^{x-a} + 1)(10^{x-b}+1).$$ To turn this into a quadratic equation, we define the substitution $y = 10^x$. Then we have $$-1/f = (10^{-a}y+1)(10^{-b}y +1)$$ and expanding this gives $$-1/f = 10^{-(a+b)}y^2 + (10^{-a} + 10^{-b})y + 1.$$ We can now apply the quadratic formula for $y$, noting that by the definition of $y$, it is positive, so we only take the positive solution (if it exists). This gives $$y = \frac{-(10^{-a} + 10^{-b}) + \sqrt{(10^{-a} +10^{-b})^2 - 4 \times 10^{-(a+b)} \times (1 + 1/f)}}{2 \times 10^{-(a+b)}}.$$ Then, we invert the substitution to get $x = \log_{10}(y)$, again noting that for certain values of $a, b$ and $f$, a solution for $x$ may not exist. For example, if $f$ is positive, as the right hand side of the initial equation is always negative, a solution will not exist. In fact, the right hand side is always between $-1$ and $0$, so $f$ needs to lie in the interval $(-1, 0)$ for there to be a solution.