Can Base 10 Logs be Evaluated Without a Calculator?

[Flexington]

Im looking for a method of evaluating large logs without the use of a calculator, such as,
log₁₀2⁷⁵⁰.

Using properties of logs i can reduce the problem to 750*10^y=2.
However this requires an estimate on the value y which is going to incure a significant error in my final result.
Im guessing I am overlooking something?

 

[tiny-tim]

Hi Flexington!

log₁₀x = lnx/ln10

log₂x = lnx/ln2

 

[HallsofIvy]

Im looking for a method of evaluating large logs without the use of a calculator, such as,
log₁₀2⁷⁵⁰.

Using properties of logs i can reduce the problem to 750*10^y=2.
However this requires an estimate on the value y which is going to incure a significant error in my final result.
Im guessing I am overlooking something?

log₁₀(2⁷⁵⁰)= 750 log₁₀(2)

 

[micromass]

Also,

$$2^{10}\sim 10^3$$,

Thus

$$\log_{10}(2^{750}) \sim \log_{10}(10^{225})=225$$

I'm curious how good this approximation actually is...

 

[George Jones]

Also,

$$2^{10}\sim 10^3$$,

Thus

$$\log_{10}(2^{750}) \sim \log_{10}(10^{225})=225$$

I'm curious how good this approximation actually is...

Very close! Maple gives 225.77

 

[Integral]

Im looking for a method of evaluating large logs without the use of a calculator, such as,
log₁₀2⁷⁵⁰.

Using properties of logs i can reduce the problem to 750*10^y=2.
However this requires an estimate on the value y which is going to incure a significant error in my final result.
Im guessing I am overlooking something?

We used to use log tables.

 

[George Jones]

We used to use log tables.

... but now tables are made out synthetic material. Sorry, couldn't resist.

 

[George Jones]

Maple gives 225.77

This was silly. From what mircromass wrote,
$$\log_{10}(2^{750}) = 225 + 75 \log_{10}(1.024)$$

 

[Flexington]

spot on micromass. Can your method be generalised?

 

[micromass]

spot on micromass. Can your method be generalised?

I don't know any generalization. The $2^{10}=10^3$-trick is something I learned from professor Daubechies. I'm guessing you could generalize it with a instead of 2, but then you would need to find suitable n and m such that $a^n=10^m$...

 

[robert2734]

If 2^10 really equaled 10^3, then the log₁₀2 would equal .3. Being an electrical engineer, I happen to know the log₁₀2 would equal .30103...


Other than memorizing more of the log table, you can either derive it from the taylor series log(1+x)=x-x^2/2+x^3/3... or interpolate from a log you do know using calculus. Say you know log(a), then the log(a+x)=log(a)+x/a+...

For example, I know the log(10)=2.3, so therefore log(11)=2.3+1/10=2.4. Close enough for engineering anyways.

 

[tiny-tim]

in shakespeare's time, one could get one's servants to do it!

from http://www.enotes.com/romeo-and-juliet-text/act-iv-scene-iv" …

SECOND SERVANT:
I have a head, sir, that will find out logs ​

 

[George Jones]

This was silly. From what mircromass wrote,
$$\log_{10}(2^{750}) = 225 + 75 \log_{10}(1.024)$$

When I wrote the above, I had in mind calculating by hand the first couple of terms from

you can either derive it from the taylor series log(1+x)=x-x^2/2+x^3/3

but then I realized that

I know the log(10)=2.3

is also needed. Given this,
$$\log_{10}(2^{750}) \doteq 225 + \frac{75}{2.3} \left(0.024 - \frac{0.024^2}{2}\right),$$
which is tedious, but not difficult, to calculate by hand, and which gives a very accurate result.

 

[Pengwuino]

Thank god for texas instruments.

 

[robert2734]

The natural log of the first twenty numbers have pretty good one digit approximations.

ln2=.7 ln3=1.1 ln4=1.4 ln5=1.6 ln6=1.8 ln7=1.95 ln8=2.1 ln9=2.2 ln10=2.3 ln11=2.4 ln12=2.5 ln13=2.6 ln14=2.65 ln15=2.7 ln16=2.8 ln17=2.85 ln18=2.9 ln19=2.95 ln20=3.0 ln(21)=3.05

If you need more accuracy, you can just use calculus.

 

[dodo]

And, of course, you can always do a binary search for more precision on one of them:

If exp(2.95) > 19 but exp(2.94) < 19, then ln(19) must be in between... oh, let's try some number in the middle...

 

[robert2734]

Say I'm sitting in an art movie and need to know 7^18. 7^18=e^35.1=(20^11)*8=2^14*10^11=16*10^14. My calculator says 1.628e15. If you need more accuracy than this get a calculator.

What's the smallest number where my method loses monotonicity? That is you'll get a smaller log from a bigger number than from some smaller number? I have ln(3125)=ln(5^5)=5*1.6=8.0. ln(3072)=ln(3*2^10)=1.1+10*.7=8.1.

The ln(5) is really 1.61 so the ln(3125) is really 8.05. The ln(2) is really .693 so ln(3072) is really 8.03.

 

[dodo]

Say I'm sitting in an art movie and need to know 7^18.

This was hilarious. (The movie must have been Aronofsky's "Pi".)

I did a silly suggestion, anyway, because for an iterative approach you'd need a calculator anyway.

I'm wondering if there is such thing as monotonicity here; results are up or down the real value mostly depending on which direction the initial 1-digit approximation was rounded.

 

[robert2734]

ln(48)=4*0.7+1.1=3.9 ln(49)=2*1.95=3.9 ln(50)=2.3+1.6=3.9.

It isn't too hard to realize that in real life ln(50)>ln(49)>ln(48). But rounded to one digit there's no incongruety. Only slightly harder to realize that they must be 1/50 or .02 apart. Anyways what I meant was where do you see an obvious error because the log of a smaller number is larger than the log of a bigger number.

It's probably worth memorizing the log of small primes to better than one digit accuracy. Because when you take 2^10, you lose a digit of precision. If you take 2^100, you lose two digits of precision.