Let f(n) be the base-10 logarithm of the sum of the elements of the \(n\)th row in Pascal's triangle. Express \(\frac{f(n)}{\log_{10} 2}\) in terms of n.
+499 Hey there guest! Glad to see you posting something about logarithms!(I need to brush up on them lol).
Let's get to the problem.
It's first given that f(n) is the base 10 logarithm of the sum of the elements in the nth row of pascals triangle.
A key thing that you should realize is that the nth row of pascals triangle can be represented as: \({n \choose 0} + {n \choose 1} + {n \choose 2}.... {n \choose n}\)
Coincidentally, this also equals \(2^n\). You can try this out yourself with the first few rows of pascal's triangle!
The function f(n) then becomes:
\(f(n) = \log_{10}2^{n}\)
The expression we are asked to simplify then turns into:
\(\log_{10}(2^n)/ \log _{10} 2\)
Here, we have a neat log property we can use:
\(\log_{a} b / \log_{a} c = \log_{c} b\)
We then convert
\(\log_{10}(2^n)/ \log _{10} 2 = \log_{2}2^n\)
With the basic definition of a log, we then have
\(\log_{2}2^n = n\)
So the expression in terms of n is just n
Quick edit / note : Remember that the "first row" of pascal's triangle with just 1 term, is actually counted as the "0th" row.
