Solving for $x$ using $\ln$ or any possible way.

Question

$$ 12.46x=1-(1+x)^{-20} $$ I tried solving for $x$ using $\ln$ and other methods but the only answer i got was 0.8. The correct answer is approximately to $0.05$.

Answer 1

The equation can be rearranged to give $$(1-12.46 x) (x+1)^{20}-1=0.$$

Expanding the $20^\text{th}$ power yields $$-12.46 x^{21}-248.2 x^{20}-2347.4 x^{19}-14014.4 x^{18}-59228.7 x^{17}-188335. x^{16}-467446. x^{15}-927139. x^{14}-1.49207\times 10^6 x^{13}-1.96681\times 10^6 x^{12}-2.1341\times 10^6 x^{11}-1.90803\times 10^6 x^{10}-1.40163\times 10^6 x^9-839929. x^8-405430. x^7-154420. x^6-44864.7 x^5-9359.4 x^4-1227.4 x^3-59.2 x^2+7.54 x=0$$

which by Descartes rule of signs has exactly one positive root.

Solving for it yields $x=0.0500209$.

Answer 2

An easier way to get DumpsterDoofus's result:

Let $f(x) =(1+ax)(1+x)^n-1 $ where $a > 0$ and $n > 0$.

$f(0) = -1$ and $f(1) =(1+a)2^n-1 > 0 $ since both $1+a$ and $2^n$ are greater than $1$, so $f$ has a root between $0$ and $1$.

$f'(x) =a(1+x)^n+(1+ax) n(1+x)^{n-1} >0 $ since both terms are positive.

Therefore $f$ has exactly one root between $0$ and $1$.