1. ## Logarithmic Functions

This problem actually does not involve calculus, but eventually we will need to take derivatives of this expressions. The instructions indicate: Use logarithmic rules(like product,power,quotient rules) to rewrite each expression in terms of log3 2 and log3 5. Here is the problem: log3 270. (The 3 is lowercase, 270 is larger). The answer is: 3+log3 2+log3 5. If someone can illustrate how to get this, I would appreciate it, Thanks for the help.

2. Originally Posted by kdogg121
This problem actually does not involve calculus, but eventually we will need to take derivatives of this expressions. The instructions indicate: Use logarithmic rules(like product,power,quotient rules) to rewrite each expression in terms of log3 2 and log3 5. Here is the problem: log3 270. (The 3 is lowercase, 270 is larger). The answer is: 3+log3 2+log3 5. If someone can illustrate how to get this, I would appreciate it, Thanks for the help.
$\displaystyle log_3(270) = log_3(2 \cdot 3^3 \cdot 5)$

$\displaystyle = log_3(2) + log_3(3^3) + log_3(5)$

$\displaystyle = log_3(2) + 3 \cdot log_3(3) + log_3(5)$

$\displaystyle = log_3(2) + 3 \cdot 1 + log_3(5)$

$\displaystyle = 3 + log_3(2) + log_3(5)$

-Dan

3= 2+5e^-4x I took the natural log (ln) for all the terms so that the e term would cancel out and I was left with ln3= ln2-4X I then isolated the x term and had ln1= -4x and got an answer of x= ln1/-4 which equals 0. The book has the answer as: ln5/4 =0.402 approx. What did I do wrong?

4. Hello,

$\displaystyle \log(a-b) \; \text{is not equal to} \; \log(a)-\log(b)$

Firstly, simplify 2 and 3.
Then you can put the logarithm, and use the following properties to isolate x :

$\displaystyle \log(ab)=\log(a)+log(b)$

$\displaystyle \log(e^a)=a$

5. I'm sorry I still don't understand how that will work out, could someone illustrate what should be done? My professor said to cancel the e by using the ln on each term. Could you show how the problem is worked out?

6. 3= 2+5e^-4x

This is the same as 1=5e^(-4x)

You can't compose a sum with the logarithm, it's a nonsens and unuseful, because you won't be able to isolate x.

So now you can compose with the logarithm.

$\displaystyle log(1)=log(5 e^{-4x})$

As mentioned above (i gave you the general formulae you needed...) :

$\displaystyle log(5 e^{-4x})=log(5)+log(e^{-4x})=log(5)-4x$

=> the equation will now be :

$\displaystyle 0=log(5)-4x$

And you have x ;-)

7. Ok, I follow what you are doing up to the end, but I think I need to have X isolated, to solve for it. What I did was subtract Log5 from the left side of the equation, which would give you -Log5= -4X I then divided both sides by -4 and the answer I got when computing this on my calculator was different than the book's answer of: .402 How do I get this?

8. Hello,

You have to differentiate ln and log (used on calculators for ln in basis 10). Otherwise, if you type ln(5)/4, you will have the correct answer :-)

9. I was wondering whether I have the correct answer for this problem:

ln of cubed root of x^2-x

I have as an answer: 1/3(2lnx+lnx) Is this correct? Thanks...