logarithmic/exponential inequalities

[civic arrow]

I don’t even understand the question

[tight slate]

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[hot flume]

The number of digits in a positive integer n is ⌊lg(n)⌋ + 1.

[civic arrow]

Wait what?

How?

They never taught us this formula

[civic arrow]

Also what about negative integers? Is it just gonna be lg(n)?

[hot flume]

Well, it's pretty easy to derive it.
If n = 10^k, then the number of its digits is k + 1, which is lg(n) + 1.
As that is the same for all numbers between 10^k and 10^(k + 1) - 1, we can just round it down, getting ⌊lg(n) + 1⌋ = ⌊lg(n)⌋ + 1.

[civic arrow]

Ok so in this particular case

I gotta find the number of digits in 4^100

The formula I’m working with

Is lg(n)+1

lg(4^100)+1

100lg4 +1

We were given the value of log 4 above

Let me see

It’s 0.60206

[hot flume]

Don't forget that it's ⌊lg(n)⌋, not just lg(n).

[civic arrow]

I solve the log first and then add 1?

[hot flume]

What I mean is that you need to round lg(n) down.

Then add 1.

[civic arrow]

So [100*0.60206]+1
(60.206)+1

Rounding 60.206 to 60

61

Is that correct sir?

[civic arrow]

Gotcha

[hot flume]

Yes.
Also, why "sir"? I'm not that old 😅

[civic arrow]

Or else the number of digits will be in decimals and that won’t make a lot of sense?

[civic arrow]

I was just trying to be respectful lol

[hot flume]

Yes.

[civic arrow]

I see also

What if

N is negative?

Asking this question just out of curiosity

[hot flume]

Well, negative numbers don't really have a different amount of digits. So, we can just replace n with |n|.

[civic arrow]

True they don’t

I see for them the formula is gonna look like [ lg|n|] + 1?

Asking this question

Before I write it

In my notes

[hot flume]

I don't recommend using [x], as it's a little ambiguous. I prefer only using ⌊x⌋ and ⌈x⌉.

[civic arrow]

Oh there are different kinds of box brackets now?

[hot flume]

Well, ⌊x⌋ is the floor function, ⌈x⌉ is the ceiling function.

[civic arrow]

Ya idk about dat

🗿

Thanks a bunch tho

I won’t waste your time anymore

Thanks🌻