Alan

Nowhere does the original question call 719 the half-life.  

The structure of the equation implies that 1/719 is the decay constant, so the half-life would be ln(2)/(1/719) or just 719*ln(2).

See:  http://web2.0calc.com/questions/m-t-85e-t-29-describes-the-amount-of-a-radioactive-isotope-as-a-function-of-time-t-years-after-how-long-the-amount-has-dropped-to-1-10-of#r9

Hmm!

Radioactive decay equation often written as:  \(N=N_0e^{-kt}\) where k is the decay constant.

When \(N=\frac{N_0}{2}\) we have \(\frac{1}{2}=e^{-k\tau}\) where \(\tau\) is the half-life.

Taking logs we get \(\ln{\frac{1}{2}}=\ln{e^{-k\tau}}\)  or \(-\ln2=-k\tau\)  so the relation between half-life and decay constant is \(\tau=\frac{\ln2}{k}\)  or \(k=\frac{\ln2}{\tau}\)

solve the equation 4x-5=11x+16

Subtract 4x from both sides:

-5 = 7x + 16

Subtract 16 from both sides:

-21 = 7x

Divide both sides by 7:

-3 = x    (or just x = -3)

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Take logs of both sides:

ln(1.6491633333)=ln((1.023)^2n)  where ln represents log to the base e

Now use the property  ln(x^a) = a*ln(x) to get;

 ln(1.6491633333)=2n*ln(1.023)

Rearrange:  n = ln(1.6491633333)/(2*ln(1.023))

You can use a calculator to evaluate this.

Only true in the limit as x tends to minus infinity!

cos(90°) = 0 Max!!

There are an infinite number of possibilities for a and b!

For example:

a = 5/13, b = 2/13

a = 1/(5 - 2*sqrt(3)), b= 0

etc.

You need another condition to establish a unique solution.

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This graph might help you answer your question

(x-2i)(x+2i)(x-3) → (x^2 + 4)(x -3) → x^3 - 3x^2 + 4x - 12