# Thread: Exponents and Logarithms-- Exponent question

1. ## Exponents and Logarithms-- Exponent question

Hi guys i just started this new unit, have a few question not having learned about logarithms, What would the formula look like?

1) It is estimated that 30% of a certain substance decays in 20 hours. what is the half life of the substance.
-- This is what i do to solve Y= AB^x- 50%=100%(100-30%=70 or .7%)^T = .5%=(.7)^T What is T in this case and how do i incorporate the 20 hours into this question? as well did i use the data given in the question correctly?

2. ## Re: Exponents and Logarithms-- Exponent question

I would set the problem up as follows:

$A(t)=A_0e^{-kt}$ where $A_0$ is the initial amount, and $t$ is measured in hours.

Now, we are told that when $t=20$ we have $A(t)=0.7A_0$ and this gives us:

$A(20)=A_0e^{-20k}=0.7A_0\,\therefore\,e^{-20k}=0.7$

Now, solve this for $k$ by converting from exponential to logarithmic form, then substitute into the original function. Then equate this to $0.5A_0$ and solve for $t$. What do you find?

3. ## Re: Exponents and Logarithms-- Exponent question

I have not learned about logarithms so i dont know what that form is. Would i need to know about logs to solve?

4. ## Re: Exponents and Logarithms-- Exponent question

I find it hard to conceive of being assigned a problem like this without first being taught about logarithms, and how to convert an equation from exponential to logarithmic form.

$\log_a(b)=c$ implies $a^c=b$

5. ## Re: Exponents and Logarithms-- Exponent question

It seems your correct my teacher assigned me problems that i could not solve, Now that i have moved on in the chapter i am starting to learn about logs.

Question though while solving exponential equations
1)Sqrt(256)^5 / Sqrt(64)^6 -- when solving that i dont get one step how to i get rid of those sqrts? my teacher does a step where the sqrts go away and are left with [(256)^1/5] / [(64)^1/6]

After that i can go from there but i dont get that step any help would be appreciated. As well thanks for your prompt replies MarkFL I hit the thanks button.