Can someone explain me the subjects Big O notation and logarithms please?
I can't understand those subjects
For example if I have a question like this: recall that logan is the power to which you need to raise a in order to obtain n.
The main rules for working with logarithms are the following:
loga(nk) = klogan loga(nm) = logan + logam nlogab = blogan logan⋅logba = logbn Is it true that (log5n)2=2log5n?
Question 2: 2.log2n⋅log32=log3n?
Questions like this... please help
Ok, I'll try to help you.
You see, when logarithmic equalities are to be proved, you are best off raising to a suitable power.
So we said that $\log_a n$ (that's the way it's written) is the power to which $a$ must be raised to get $n$. In other words , $a^{\log_a n} = n$.
Now, we'll see how we translate from logarithms to powers, and make working very comfortable.
To show that $\log_a (n^k) = k \log_a n$, we do this: let $b=\log_a (n^k)$,then $a^b = a ^ {\log_a (n^k)} = n^k = (a^{\log_a n})^k = a^{k\log_a n}$. Since the powers have to be the same, $b=k\log_a n$.
I'll do one more and leave the rest as exercises. Say you want to show that $\log_a n \cdot \log_b a = \log_b n$. Let $c=\log_b n$. Then note that $b^c=n = a^{\log_a n} = (b^{\log_b a})^{\log_a n} = b^{\log_a n \cdot \log_b a}$. Now the exponents have to be the same,hence $c=\log_a n \cdot \log_b a$. That's all, it's child's play, and the rest of the exercises, including question $2$ will come under this.