# Thread: Exponential Equation & Decay

1. ## Exponential Equation & Decay

Hello there! I was just wondering if someone could kindly show me how to do the following questions:

1) 4^x + 6(4^-x) = 5

2) The half-life of a certain substance is 3.6 days. How long will it take for 20g of the substance to decay to 7g?
-The only equation I know of is y = ab^x but I'm not sure what to put where, thank you!!

2. Hint: use logarithms: $log (x^y) = y log x$

Hello there! I was just wondering if someone could kindly show me how to do the following questions:

1) 4^x + 6(4^-x) = 5

2) The half-life of a certain substance is 3.6 days. How long will it take for 20g of the substance to decay to 7g?
-The only equation I know of is y = ab^x but I'm not sure what to put where, thank you!!
$4^x + 6(4^{-x}) = 5$

multiply every term by $4^x$ ...

$4^{2x} + 6 = 5(4^x)$

$4^{2x} - 5(4^x) + 6 = 0$

$(4^x - 3)(4^x - 2) = 0$

$4^x = 3$ ... $4^x = 2$

$x = \frac{\log{3}}{\log{4}}$ ... $x = \frac{1}{2}$

$y = 20\left(\frac{1}{2}\right)^{\frac{t}{3.6}}
$

set $y = 7$ and solve for $t$

2) The half-life of a certain substance is 3.6 days.
How long will it take for 20g of the substance to decay to 7g?

The only equation I know of is: . $y \:=\: ab^x$
but I'm not sure what to put where, thank you!
You can derive the half-life formula yourself . . .

The general form is: . $A \:=\:A_oe^{-kt}$ .[1]

. . where: . $\begin{array}{ccc}A &=& \text{amount at time }t \\ A_o &=& \text{original amount} \\ k &=& \text{constant} \end{array}$

We are given: . $\text{amount }\tfrac{1}{2}A_o\text{ at }t = 3.6$

Substitute into [1]: . $\tfrac{1}{2}A_o \:=\:A_oe^{-3.6k} \quad\Rightarrow\quad e^{-3.6k} \:=\:\frac{1}{2}$

Take logs: . $\ln\left(e^{-3.6k}\right) \:=\:\ln\left(\tfrac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \:=\:-\ln2$

. . Then: . $-3.6k \:=\:-\ln 2 \quad\Rightarrow\quad k \:=\:\frac{\ln2}{3.6}$

Hence, the function is: . $A \:=\:A_oe^{-\frac{\ln2}{3.6}t}$

For this problem, $A_o = 20\quad\Rightarrow\quad A \:=\:20e^{-\frac{\ln2}{3.6}t}$

Now we can answer the question: When will $A = 7$ ?

We have: . $7 \:=\:20e^{-\frac{\ln2}{3.6}t} \quad\Rightarrow\quad e^{-\frac{\ln2}{3.6}t} \:=\:0.35 \quad\Rightarrow\quad -\frac{\ln2}{3.6}t \:=\:\ln(0.35)$

. . $t \:=\:-\frac{3.6}{\ln2}\!\cdot\!\ln(0.35) \:=\:5.452463422 \;\approx\;5.45$ days

5. Originally Posted by Soroban

You can derive the half-life formula yourself . . .

The general form is: . $A \:=\:A_oe^{-kt}$ .[1]

. . where: . $\begin{array}{ccc}A &=& \text{amount at time }t \\ A_o &=& \text{original amount} \\ k &=& \text{constant} \end{array}$

We are given: . $\text{amount }\tfrac{1}{2}A_o\text{ at }t = 3.6$

Substitute into [1]: . $\tfrac{1}{2}A_o \:=\:A_oe^{-3.6k} \quad\Rightarrow\quad e^{-3.6k} \:=\:\frac{1}{2}$

Take logs: . $\ln\left(e^{-3.6k}\right) \:=\:\ln\left(\tfrac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \:=\:-\ln2$

. . Then: . $-3.6k \:=\:-\ln 2 \quad\Rightarrow\quad k \:=\:\frac{\ln2}{3.6}$

Hence, the function is: . $A \:=\:A_oe^{-\frac{\ln2}{3.6}t}$

For this problem, $A_o = 20\quad\Rightarrow\quad A \:=\:20e^{-\frac{\ln2}{3.6}t}$

Now we can answer the question: When will $A = 7$ ?

We have: . $7 \:=\:20e^{-\frac{\ln2}{3.6}t} \quad\Rightarrow\quad e^{-\frac{\ln2}{3.6}t} \:=\:0.35 \quad\Rightarrow\quad -\frac{\ln2}{3.6}t \:=\:\ln(0.35)$

. . $t \:=\:-\frac{3.6}{\ln2}\!\cdot\!\ln(0.35) \:=\:5.452463422 \;\approx\;5.45$ days
Hello Sorobon! Wow, that certainly is a new equation for me. Thank you for the step by step instructions, I should be able to get it!

Another way to do this is to recognize that "half" life means that the amount is multiplied by 1/2 every 3.6 days- In T days, there are T/3.6 periods of 3.6 days each which means the initial amount is multiplied by 1/2 T/3.6 times: $x= 20(1/2)^{T/3.6}$. You want to solve $7= 20(1/2)^{T/3.6}$ or, dividing both sides by 20, $\frac{7}{20}= (1/2)^{T/3.6}$. To solve that take the logarithm of both sides (it doesn't matter whether you take the natural logarithm or common logarithm) $log(\frac{7}{20})= (T/3.6)log(1/2)$ so [tex]T= 3.6\frac{log(7/20)}{log(1/2)}= 3.6\frac{log(20)- log(7)}{log(2)}, exactly what Soroban got.
Another way to do this is to recognize that "half" life means that the amount is multiplied by 1/2 every 3.6 days- In T days, there are T/3.6 periods of 3.6 days each which means the initial amount is multiplied by 1/2 T/3.6 times: $x= 20(1/2)^{T/3.6}$. You want to solve $7= 20(1/2)^{T/3.6}$ or, dividing both sides by 20, $\frac{7}{20}= (1/2)^{T/3.6}$. To solve that take the logarithm of both sides (it doesn't matter whether you take the natural logarithm or common logarithm) $log(\frac{7}{20})= (T/3.6)log(1/2)$ so [tex]T= 3.6\frac{log(7/20)}{log(1/2)}= 3.6\frac{log(20)- log(7)}{log(2)}, exactly what Soroban got.