Question

A new radioactive element is found on Mars. Every 10 years, the concentration of the element is a quarter of what it was at the start of the 10-year period. If the decay rate is given by Equation with left side as, C= Co (x)^t/10, where C0 is the initial concentration of the element, C is the final concentration of the element, and t is the time in years, what is the value of x?

Answer 1

Answer:

$\boxed{\dfrac{1}{16}}$

Explanation:

$C = C_{0}(x)^{\frac{t}{10}}$

Data:

t = 5 yr

C = ¼C₀

Calculation:

$\begin{array}{rcl}\frac{1}{4}C_{0} & = & C_{0}(x)^{\frac{5}{10}}\\\frac{1}{4} & = & x^{\frac{1}{2}}\\x & = & \mathbf{\frac{1}{16}}\\\end{array}$

$\text{The value of x is \boxed{\mathbf{\dfrac{1}{16}}} }$

Answer 2

Answer:

x=0.25

Explanation:

The equation is

$C = Co*x^{\frac{t}{10} }$

For this problem, C would be equal to 0.25 * Co when t is equal to 10 years. If we put this information in the equation we're left with:

$0.25*Co=Co*x^{\frac{10}{10} }$

Now we solve for x:

$0.25*Co=Co*x^{\frac{10}{10} }\\0.25=x^{1 }\\0.25=x$

So the value of x is 0.25