Solving for x in an exponential

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To solve the inequality e^-(xt) <= y with t = 10 and y = 10^-6, the equation simplifies to e^(-10x) <= 10^-6. The correct approach involves taking the natural logarithm of both sides, leading to -10x <= ln(10^-6). After correcting for a negative sign, the solution for x is calculated as x = ln(10^-6) / -10, which yields approximately 13.8. Verifying this value by substituting it back into the original equation confirms its accuracy.
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Homework Statement


My problem is: e^-(xt) <= y, where t = 10 and y = 10^-6

So: e^-(x10) <= 10^-6

I have to find a value x that would make the probabilty less than or equal to 10^-6.

Homework Equations





The Attempt at a Solution


I am not sure but my attempt in finding x is:

x*10 =ln(10^-6)
x=ln(10^-6) / 10

I am not so sure that is right. I'm having somewhat of difficulty treating the e in the problem. Any help is greatly appreciated. Thanks
 
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Thats right in taking natural logs of both sides to start with, but you seem to have missed a negative sign in your first step.
 
Ok, so I would get ln(-x*10) <= ln 10^-6
x = ln(10^-6) / -10)
If I do that I calculate x to be 13.8. Does this seem right?
Put x= 13.8 back into the original equation. Does it satisfy the equation?
 
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In problems like this, you can always confirm your answer by going back to the original equation. Since you've found x and you know t, just calculate e-xt and see if it gives the answer you want.
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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