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.
What about this one: We have to solve the given equation for x.
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?
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.
As mentioned above (i gave you the general formulae you needed...) :
=> the equation will now be :
And you have x ;-)
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?