Find the exact value of the logarithmic expression. log36 C. 36 D. 12 26

Given,

Fossilized leaf contain 28% of its normal amount of carbon -14 

Half life of carbon -14 = 5600 years

We have to find the how old the fossil is .

Let's assume

The normal amount of  carbon -14  =  N

Decay constant = k 

Half life of  carbon -14  =  $\frac{0.693}{k} = 5600$ 

So,

$k= \frac{0.693}{5600}$

Leaf contain the amount of carbon - 14 = n =  $N\times \frac{28}{100} = 0.28N$ 

We have the decay equation ; $n =N{e}^{-kt}$ 

Where t is the time in years

Substituting the values in the above equation

$0.28N = N{e}^{\frac{-0.693t}{5600}}$

$0.28 = e^{\frac{-0.693t}{5600}}$ 

Taking logarithm both sides

$\ln (0.28) = \ln e^{-\frac{0.693}{5600}t}$   

$-1.273 = -\frac{0.693}{5600}t$                         $\left(because \ln a^b = b\ln a\right)$

$t = \frac{5600\times 1.273}{0.693}$

$t = 10284$

So,

The correct option is B                                             Answer