Discovering the Half Life of a Radioactive Substance: A Step-by-Step Guide

[Amel]

Homework Statement

Find the half life (in hours) of a radioactive substance that is reduced by 5% in 65 hours.

Homework Equations

The Attempt at a Solution

So looking around I find that A=A₀e^-kt

but I am not really sure how to solve this without having an initial ammount.

 

[zcd]

You don't need an initial amount. You're given the rate at which it decays, so mathematically:
$$A=A_{0}e^{0k}$$
$$0.95A=A_{0}e^{65k}$$

 

[Amel]

never mind figured it out. but if anyone xould help me wih this one it would be greatly appreciated.

Solve for x: 4^x=e^{x + 1}

how do you solve for x when its on both sides. How ever I did it the x would cancel.

 

[Amel]

Didn't see your post but thanks for the help. I used Q=Q₀(1/2)^(t/T) and it worked out for all my half-life problems.

 

[Dick]

never mind figured it out. but if anyone xould help me wih this one it would be greatly appreciated.

Solve for x: 4^x=e^{x + 1}

how do you solve for x when its on both sides. How ever I did it the x would cancel.

Take the log of both sides.

 

[Amel]

ok this is what I am doing

log of both sides, i get

(X) log(4) = (x + 1) log(e)

.60X = (X + 1) .43

I don't know how to get both Xs to one side. I have done this with natural log as well and get

X ln (4) = X+1

but how do I move the X

I even tried doing

4^X = e^xe¹

If someone could walk me through this one please, I have been trying so much crap my algebra is getting all messed up.

 

[Elucidus]

ok this is what I am doing

log of both sides, i get

(X) log(4) = (x + 1) log(e)

.60X = (X + 1) .43

I don't know how to get both Xs to one side. I have done this with natural log as well and get

X ln (4) = X+1

but how do I move the X?

How would you solve

$$ax = x + 1\; ?$$

Once you've figured that out, just let a = ln(4).

--Elucidus

 

[mbisCool]

Start from xln(4)=x+1. Are you familiar with subtraction :D? What could you subtract from the right that would bring all x's to the left?

 

[Amel]

ok so X is 2.588 right?

thats what I was doing the whole time as well but for some reason I would see x-x instead of 1.38X-X lol thank you. I know I was missing something. I hate when I am doing things this late and my mind is just not working right. on the last few I would not be able to do anything and then the next morning when its already to late (due at 5 AM online) the answers would come to me.

 

[Amel]

Ok now I can't get Solve for X : 4e^6x=6e^7x

then get 6x ln(4e) = 7x ln(6e)

6x (2.38) = 7x (2.79)

Now how do I get Xs on the same side without canceling them? if I divide then the Xs cancel.

 

[HallsofIvy]

This is now the same as Ax= Bx. How would you solve that? Seriously, if you can't solve equations like that, you need to go back and review basic algebra before you try problems involving exponentials and logarithms.

 

[Elucidus]

Ok now I can't get Solve for X : 4e^6x=6e^7x

then get 6x ln(4e) = 7x ln(6e)

6x (2.38) = 7x (2.79)

Now how do I get Xs on the same side without canceling them? if I divide then the Xs cancel.

Be careful! You are abusing the power rule of logarithms here.

The natural log of 4e^6x is not 6x (ln 4e). It is ln(4) + 6x.

--Elucidus