# Clever Logarithm Problem

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.

Solved the problem in… 7 seconds 😂

I solved these types in class 8

done.

i solved it without using a pen or change the base.

i have an excell and i use it.

the result is 7

Why didn't we solve this problem by log?

This one is for you and your ghost goats presh https://youtu.be/LS_jzDbfccE

7

Use log₂ throughout. It's cleaner and quicker, and you can just do the whole thing in your head.

8

I solved it in less than 10 seconds, the most common question of logarithms

People should stop laughing about how easy it was and just keep supporting the channel. If they think this channel has viewers who are middle-aged men like them then they are mistaken. This channel has all kinds of viewers. Some are just interested in maths regardless of their age. So should they feel down or give up after looking at the difference in skill between them and the people in the comments? This is a repetitive pattern I've been seeing in a lot of Presh's videos lately and so I just wanted to bring it up. No offense to any of you who find it easy. I understand this might seem easy to you but for someone who hasn't still learned about logarithms, this is nonetheless very interesting. If these questions are so easy for you then please don't click on the video after seeing the thumbnail if all you're going to do is just boast about your intellect.

え？日本の高校ではこれは公式として暗記させられます。🇯🇵

Bingo solved within 5 seconds…answer is 7..thanks presh I thought I would never be able to solve ur questions…

I gave a math teacher a problem based on this formula

Saw notification

Me : hmm 7?

7

It is a very easy question…

If you know a little bit logarithm rules you can solve it in 7 seconds without using a pen

LOL! Even a kid can solve it..

WOOOOOOW😍😍👏👏👏

too ez

well.. i actually got this one right and fast. bravo me finally

This time you kept it too simple man!!

Solved under 10 seconds in my head..

I converted to log base 2 instead. Nevertheless, the result is the same.

At the beginning I saw 2 and 128 and remembered that 2^7 = 128 and 7 was the answer!!!

But then i solved it.

Solved without even clicking the video!

Solved it just when it popped up in notifications.😅😅

'Solve in seconds'

Yup, about a million seconds

This is actually beggining of logarithma

Aren’t logs amazing how they knock each other out and simplify the problem!

Telescopic product and its subsidiaries are often used in mathematical Olympiads and is a crucial concept…

Question was easy but concept was quintessential..

I misread the equation as log 2 × log 3 × log 4 × … × log 128

7

I did it orally!

So easy.

Not according to the standards of this channel

easy

I solved it before even watching it (from thumnail)

Yayyyy

I could tell that this was going to be one of those telescoping problems, so in the end it would have to be log 128/log 2 which is clearly 7. It's always cool when something that appears to be rather ugly looking has an elegant solution

It is tooooooooo easy lol

When did your balls drop? Your voice started so deep xD

I do questions like this in my dreams

I was literally thinking of cancelling by multiplication between logs before yesterday!

Before watching the video – I say the answer is 7.

I also run a YouTube channel on knowledge, you will surely like it.

Enjoy!

I solve this in less than 10 seconds

1:54.

Telescoping product – nice term!

Another good video to freshen our concepts.

Thanks!

You didn't mention any diverse mathematicians this time, when solving this really easy problem that no one is going to have to think about.

Why use the natural logarythm? It doesnt matter which one you use

7

I literally solved this in 4 seconds

It's just log base 2 of 128

Presh,I love your videos,but you are probably overusing telescopic series.

There are also easier method knowing two formulas. K * log(x) = log(x^k) and a^(log_a(b)) = b. So you will get in final log_2(128).

The thought at 0:36 is known Mr.Presh please display new ones

Скажу больше, есть формула смены чисел: log a(b) * log c(d)= log a(d) * log c(b). Благодаря этой формуле, я устно передвинул все элементы вправо, а последнее число 128 поставил на первый логарифм. И результат тот же!

İ Find 7 😎

Solve this problem 2xsquare+4x-8=9

Who else beleives he would probably not solve it if he didn't see the telescope?

Wow… Wonderful… Many problems of life are appearing complicated yet simple for those who dares and works

How i can sent my questions? The question is a problem, i tried but it too hart for me. I asked my teacher but she forget.

I actually used the power rule for logs (log(x^n) = n*log(x)) and the identity that n^log_n(x) = x to collapse them all in, starting with log_2(3)*log_3(4) = log_2(3^(log_3(4))) = log_2(4), then doing the same for log_4(5) to get log_2(5), etc., all the way up to log_2(128) = 7, no base changes of any kind required

00:40 'recall the…' oh crap, I'm done. Just show me the answer. =D

Finally a problem that can be solved by me in MYD

that was easy

Please also make video on centres in circles in your style

7 ?

It actually can be solved even easier.

log_a(b)•log_b(c)=log_a(c) is where change of base formula comes from. Proof is trivial as well. Consider this:

c=b^log_b(c)=(a^log_a(b))^log_b(c)=a^(log_a(b)•log_b(c))

On the other hand:

c=a^log_a(c)

Therefore:

a^(log_a(b)•log_b(c))=a^log_a(c) and

log_a(b)•log_b(c)=log_a(c). And from this you derive change of base formula by dividing both sides by log_a(b). So this question already is a telescoping product, no need to use anything else

EZ question, the answer is 7! Solved it in 2 seconds

Easyyyyy

We are saved it is not an indian who discovered the logarithms.

Presh would have released something like the gougu theorem …

(f°f)(x)=x²; f:R-R ;

Can anyone give an example of function with this property?And how many functions with this property exist?

I have no idea why this formula was never taught to me in school. It seems so easy

Yay I solved one

7

It is such an easy problem I've ever seen

7

It's not clever it's straightforward 🙁

Amazing

Why on earth would you need to use natural logarithms just to revert to log base 2 at the end?

I remembered change of base, but not telescoping. Crud.

Easiest question u solve

Most simple .. wasn't expecting this kindda stuff from u man

Let log_a(b) = x and log_b(c) = y. Then a^x = b and b^y = c. Thus, (a^x)^y = c. So a^(xy) = c. But by the definition of logs, this means log_a(c) = xy = log_a(b) * log_b(c). The result follows from repeated application of this general principle.

Can you give a bit hard question

Next time

As this was super easy

I think i legit did this in under 5 seconds

Ye kuch jyada hi asan tha ?

7

I hate this channel,so gr8 at making me feel like im a failure.

Bhai ye to jee main se bhi easy question tha😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂

Is it true that Mcdonalds' McRib is back for a limited time?

7

I changed all logarithms to Base 2

Great channel!

Solved it in like 3 seconds 😂😂

I too solved it at once.

7

Simply use log(a)b *log(b)c = log(a)c rule 125 times 🙂

It can be easily solved without changing the base using log(y)*x= log(y^x) and i think its way easier that way.

Easy