Wolfram|Alpha returns the wrong result: how can I solve this "high precision" equation? $$1-(1-1.40*10^{-36})^x \ge 1.09*10^{-9}$$
I want to estimate $x$ such that the probability on the left becomes larger than the probability on the right.
A solution must exist because $1-(1-1.40*10^{-36})^0=0$, $\lim\limits_{n \to \infty} 1-(1-1.40*10^{-36})^n=1$ and the function is continuous.
However Wolfram|Alpha seems to return a wrong result.
Is my reasoning correct? Is there any (practical) way to estimate the result? For my problem would be sufficient to find the order of magnitude of $x$.
 A: You can solve the inequality by logarithming :
We get $1-1.09*10^{-9}\ge (1-1.4*10^{-36})^x$
Taking logarithm gives
 $$\frac {\log(1-1.09*10^{-9})}{\log(1-1.4*10^{-36})}\le x$$
Note that we divided by a negative number!
GP/PARI gives 
$$\frac{log(1-1.09*10^{-9})}{log(1-1.4*10^{-36})}\ =$$
$$778571428995750000308340238.34730623955575580298358$$
So, $x$ must be very large
An excellent approximation of the given value is $\frac {1.09*10^{-9}}{1.4*10^{-36}}$ because for $x \approx 0$ we have $log(1+x)\approx x$. And this is very easy to calculate.
A: Mathematica input:

Solve[1 - (1 - 7/(5 10^36))^x == 109/(100 10^9), x, Reals]

Result:

$$x\to \frac{11 \log (2)-2 \log (3)+11 \log (5)-\log (11)-\log (19)-\log (4441)-\log (11971)}{36 \log (2)+37 \log (5)-\log (170207)-\log (29375995111834413390753611778599)} \\
\approx 7.7857142899575*10^{26}$$

A: Could you try replacing $1.4$ by $14/10$ and $1.09$ by $109/100$ ? This would make a significant difference (I hope).
Doing the same as user111187, the solution of the equation is $$x=\frac{\log \left(\frac{100000000000}{99999999891}\right)}{\log
   \left(\frac{5000000000000000000000000000000000000}{4999999999999999999999999999999
   999993}\right)}$$ which is $$x=\frac{\log(1-109 \times 10^{-11})}{\log(1-14 \times 10^{-37})}\simeq \frac{109\times 10^{-11}} {14 \times 10^{-37}}=\frac{109}{14}10^{26}$$
Added later to this answer
The problem you posted was "simple" because it was possible to isolate the solution of the equation; it was a very nice illustration of the power of Taylor series.
But, let me take a more complex problem. What is $x$ such that $$e^{x^4}-\cos \left(x^3\right) \gt \epsilon$$ You basically could not answer this question since there is no way to find the solution of the equation. Then use Taylor to find that $$e^{x^4}-\cos \left(x^3\right) \simeq x^4+\frac{x^6}{2}+\frac{x^8}{2}+O\left(x^9\right)$$ So $x\gt \sqrt[4]\epsilon$ is the solution
