element x is a radioactive isotope such that every 18 years, its mass decreases by half. given that the…

Explanation:

Step1: Write the radioactive - decay formula

The general formula for radioactive decay is $M = M_0\left(\frac{1}{2}\right)^{\frac{t}{h}}$, where $M$ is the final mass, $M_0$ is the initial mass, $t$ is the time elapsed, and $h$ is the half - life. Here, $M_0 = 40$ grams, $h = 18$ years, and $M = 24$ grams. So the equation becomes $24=40\left(\frac{1}{2}\right)^{\frac{t}{18}}$.

Step2: Simplify the equation

First, divide both sides of the equation by 40: $\frac{24}{40}=\left(\frac{1}{2}\right)^{\frac{t}{18}}$. Simplify $\frac{24}{40}$ to $\frac{3}{5}$, so $\frac{3}{5}=\left(\frac{1}{2}\right)^{\frac{t}{18}}$.

Step3: Take the natural logarithm of both sides

$\ln\left(\frac{3}{5}\right)=\ln\left(\left(\frac{1}{2}\right)^{\frac{t}{18}}\right)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln\left(\frac{3}{5}\right)=\frac{t}{18}\ln\left(\frac{1}{2}\right)$.

Step4: Solve for $t$

We know that $\ln\left(\frac{3}{5}\right)=\ln(3)-\ln(5)\approx1.0986 - 1.6094=- 0.5108$ and $\ln\left(\frac{1}{2}\right)=-\ln(2)\approx - 0.6931$. Then $t = 18\times\frac{\ln\left(\frac{3}{5}\right)}{\ln\left(\frac{1}{2}\right)}$. Substitute the values of the logarithms: $t = 18\times\frac{-0.5108}{-0.6931}$. $t=\frac{9.1944}{0.6931}\approx13.3$.

Answer:

$13.3$ years