Here's the question you clicked on:
algebra2sucks
Show why the log100 + log10000 = 2log(3)27
compute all the numbers and see that they are equal
you need \(\log(100)\) and also \(\log(10000)\) first
if you have a power of ten , the log counts the zeros
lol i have no idea how to do this
When log 100 = a, this means: \[10^a=100\]Now, because 100 is a power of 10, you can see very quickly what log 100 is... Same for log 10000 >> what power of 10 is it?
If you have problems understanding logarithms, always remember one thing: LOGARITHMS are EXPONENTS
In your problem, I also see a logarithm with base 3. I will call it b: \[\log_{3} 27=b\]So b is that logarithm, base 3 of 27. This means, because logarithms are exponents, \[3^b=27\]Now, 27, doesn't it sound familiar? Yes! It is a power of 3! In fact, it is 3³:\[3^b=3^3\]so b = 3. The right hand side of the problem doubles that number, so we have 6 there. Can you see now, why the left hand side is 6 as well?
Yea i guess so how do we answer this
Begin with the left hand side, replace log(100) and log(10000) with the numbers we've got. These will add up to 6. Right hand side is 2*3=6 also. Now you have shown that they are equal...
so 6 is the answeR?
If you re-read the question, you'll see that "6" ids not the answer. You have to show that the left hand side and right hand side are equal. Although they seem very different, both sides give you 6 if you calculate them. Therefore they are equal. That is what you had to prove.
oh so how to i prove that with the equation right? i have no idea how to do this at all i know one side equals 6
how do we get 6 for the left hand side?
\(\log(100)\) measn \(\log_{10}(100)\) and since \(100=10^2\) we know \(\log(100)=2\)
last job is to solve \(\log(10000)\) i.e. solve \(10^y=10000\) for \(y\)
3 would be the y?
What makes you think it is 3?
because it has to equal to six idk i'm confused