Mind your decisions, I’m Presh Talwalkar. Simplify the following expression: log base 2 of 3, multiplied by log base 3 of 4, multiplied by log base 4 of 5, and so on, until the final term, log base 127 of 128. I adapted this problem from a post I saw on Reddit homework help. Pause if you’d like to give this problem a try, and when you’re ready, keep watching to learn how to solve this problem. Recall the change of base formula for logarithms: Log base b of u is equal to log base a of u divided by log base a of b. This is for u greater than zero (0) and a and b positive real numbers that are not equal to one (1). For the special case where a is equal to e, we simplify this to the natural log of u (ln u) over the natural log of b (ln b). We’ll use this version of the formula to simplify this product. The first term becomes the natural log of 3 (ln 3) over the natural log of 2 (ln 2). The second term becomes the natural log of 4 over the natural log of 3. We continue to apply this formula all the way to the very end of this product. We then notice an interesting pattern. Natural log of 3 (ln 3) in the numerator of the first term will cancel with natural log of 3 in the denominator of the second term. Then the natural log of 4 will cancel out between the second and third terms, and this pattern will continue all the way to the very final term. The pattern is reminiscent of a retracting telescope where only the very beginning and the very end survived. For this reason it’s known as a telescoping product: only the very beginning and the very end survived. We can then simplify this formula one more time using the change of base formula. The natural log of 128 over the natural log of 2 will be equal to log base 2 of 128. 128 is equal to 2 to the power 7, and therefore this simplifies to be 7, and that’s our answer. I love how this problem miraculously simplifies. Thanks for making Mind Your Decisions one of the best channels on YouTube. As always, thanks for watching and thanks for your support.