I am trying to solve for m in the following equality $$a\left(\frac{1-(1+c)^{-(m+d)}}{c}\right)(1+c)=b\left(\frac{1-(1+c)^{-m}}{c}\right)(1+c)$$ I have worked it to the following $$m = -log_{1+c}\left(\frac{b-b(1+c)^{-m}}{a}+1\right)$$
Thanks for any help.
Thanks to Moti's suggestion, I have been able to solve the equation.
Original Equation:
$$a\left(\frac{1-(1+c)^{-(m+d)}}{c}\right)(1+c)=b\left(\frac{1-(1+c)^{-m}}{c}\right)(1+c)$$
Apply property of exponents: $a^{x+y} = a^x\times a^y$
$$a\left(\frac{1-(1+c)^{-m}(1+c)^{-d}}{c}\right)(1+c)=b\left(\frac{1-(1+c)^{-m}}{c}\right)(1+c)$$
Divided both sides by $(1+c)$
$$a\left(\frac{1-(1+c)^{-m}(1+c)^{-d}}{c}\right)=b\left(\frac{1-(1+c)^{-m}}{c}\right)$$
Distributed $a$ and $b$
$$\frac{a-a(1+c)^{-m}(1+c)^{-d}}{c}=\frac{b-b(1+c)^{-m}}{c}$$
Multiplied both sides by $c$
$$a-a(1+c)^{-m}(1+c)^{-d}=b-b(1+c)^{-m}$$
Subtracted $a$ from both sides
$$a(1+c)^{-m}(1+c)^{-d}=b-b(1+c)^{-m}-a$$
Added $b(1+c)^{-m}$ to both sides
$$a(1+c)^{-m}(1+c)^{-d}+b(1+c)^{-m}=b-a$$
Factored out $(1+c)^{-m}$
$$(1+c)^{-m}\left(b-a(1+c)^{-d}\right)=b-a$$
Divided both sides by $b-a(1+c)^{-d}$
$$(1+c)^{-m}=\frac{b-a}{b-a(1+c)^{-d}}$$
Applied the natural $\log$ to both sides
$$\ln\left((1+c)^{-m}\right)=\ln\left(\frac{b-a}{b-a(1+c)^{-d}}\right)$$
Applied log rule: $\log_a(x^b)=b\times\log_a(x)$
$$(-m)\ln(1+c)=\ln\left(\frac{b-a}{b-a(1+c)^{-d}}\right)$$
Divided both sides by $\ln(1+c)$ and multiplied by $-1$
$$m=-\frac{\ln\left(\frac{b-a}{b-a(1+c)^{-d}}\right)}{\ln(1+c)}$$