Help Solving a logarithmic equation $P\times\log{(1-\frac{a}{nP})} = -b\times\log{(1+\frac{c}{n})}$ for P where P>0

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I have tried using algebraic Logarithm and exponent rules but I cannot get P into a common form. I get P in exponent and standard form or I get P in Logarithmic and standard form

My attempt so far:

$10^{P\times\log{(1-\frac{a}{nP})}}=10^{-b\times\log{(1+\frac{c}{n})}}$

$10^{P}10^{\log{(1-\frac{a}{nP})}}=10^{-b}10^{\log{(1+\frac{c}{n})}}$

$10^{P}(1-\frac{a}{nP})=10^{-b}(1+\frac{c}{n})$

$10^{P}-\frac{a10^{P}}{nP}=10^{-b}+\frac{c10^{-b}}{n}$

From here it really looks like the product log function could apply but I cannot get the equation into an $a^{x}+x=b$ form

From here I am genuinely lost as I've gone in a circle.

$nP10^{P}-a10^{P}=nP10^{-b}+cP10^{-b}$

$nP10^{P}-a10^{P}=P(n10^{-b}+c10^{-b})$

$10^{P}(nP-a)=P(n10^{-b}+c10^{-b})$

$10^{P}(nP-a)=P(n10^{-b}+c10^{-b})$