The mass m(t) remaining after t days from a 70 g sample of thorium-234 is given by the equation shown below.

m(t) = 70e-0.0274t

Find the half-life of thorium-234. (Round your answer to the nearest tenth.)

you would be solving

35 = 70 e^-.0274t
.5 = e^-.0274t
-.0274t = ln(.5)
t = ln(.5)/-.0274
= 25.297

half life is appr 25.3 years

To find the half-life of thorium-234, we need to determine the time it takes for the mass to decrease by half.

Let's denote the initial mass as m₀, and the half-life as T.

We know that the remaining mass after time T is equal to half of the initial mass:
m(T) = m₀/2

In this case, the initial mass is given as 70 g. So, we can express our equation as:

70e^(-0.0274T) = 70/2

Dividing both sides of the equation by 70, we get:

e^(-0.0274T) = 1/2

Next, we take the natural logarithm (ln) of both sides to isolate the exponent:

ln(e^(-0.0274T)) = ln(1/2)

The natural logarithm and the exponential function cancel each other out, leaving us with:

-0.0274T = ln(1/2)

Now, we can solve for T by dividing both sides of the equation by -0.0274:

T = ln(1/2)/(-0.0274)

Using a calculator, we can evaluate this expression to find the value of T. (Note: ln(1/2) is approximately -0.6931)

T ≈ -0.6931 / (-0.0274)

T ≈ 25.3

Therefore, the half-life of thorium-234 is approximately 25.3 days.