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how can i find 2^x = 3 without calculator

Anonymous Sat 24 Apr 10:06:01 2021 No.13024018 View Reply Original Report

Quoted By: >>13024036 >>13024040 >>13024060 >>13024173

i find this but it not help.
https://math.stackexchange.com/questions/1653374/easy-way-to-compute-logarithms-without-a-calculator

Anonymous

Anonymous Sat 24 Apr 2021 10:14:07 No.13024036 Report

Quoted By:

>>13024018
you can try to figure out what power of 2 is closest to 3,000,000 which is approximately 3x2^20.

Then divide by 2^10.

Anonymous

Anonymous Sat 24 Apr 2021 10:15:08 No.13024040 Report

Quoted By:

>>13024018
$ln(x)={lim}\limits_{x->\infty} n(x^{1/n}-1)$

Anonymous

Anonymous Sat 24 Apr 2021 10:19:21 No.13024051 Report

Quoted By: >>13024095

log(1+x)=x - (x^2)/2 + (x^3) /3 - (x^4)/4... Just take the first few term to have a good approximation if x is kinda small

Anonymous

Anonymous Sat 24 Apr 2021 10:21:40 No.13024060 Report

Quoted By:

>>13024018
$x=\frac{log_{b}2}{log_{b}3}$
Choose whatever b you like.

Anonymous

Anonymous Sat 24 Apr 2021 10:28:18 No.13024085 Report

Quoted By:

sorry i try to understand but i dont know what those mean.
can you explain me with like algebra or what i must study before understand this explanations.
sorry if this is stupid.

Anonymous

Anonymous Sat 24 Apr 2021 10:31:28 No.13024095 Report

Quoted By: >>13024209 >>13024217

>>13024051
Jesus christ, no; that sum converges extremely slowly

Anonymous

Anonymous Sat 24 Apr 2021 10:38:55 No.13024114 Report

Quoted By:

just solve 2^x -3 = 0 for x

Anonymous

Anonymous Sat 24 Apr 2021 10:58:27 No.13024173 Report

Quoted By:

>>13024018
It's log base 2 of 3

which is equal to ln (3) / ln(x)

ln (1+x) = x - x^2 /2 + x^ 3/3 - x^4 /4 ... The other way is to use a log table.

Anonymous

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Anonymous Sat 24 Apr 2021 11:07:11 No.13024209 Report

Quoted By:

>>13024095
Not if x is small,

| x^n / n - x^n+1 /n+1 + .... | < x^n/(n(1-x)) which quickly goes to 0
and you approximate large n by fractions. For example you write 2 = (3/2) * (2/3) so
log(2)=log(3/2) + log(2/3) = log(1 + 1/2) + log( 1 - 1/3) both of which are easy to calculate. Both take only a few terms in the power series to calculate very accurately.
For 3 you can write 3= 2 * (3/2) so
log(3)=log(2) + log(3/2) and we've already calculated both of these before.
So you can easily calculate log(3)/log(2) which is what OP wants.

Anonymous

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Anonymous Sat 24 Apr 2021 11:09:03 No.13024217 Report

Quoted By:

>>13024095
Not if x is small,

| x^n / n - x^n+1 /n+1 + .... | < x^n/(n(1-x)) which quickly goes to 0
and you approximate large n by fractions. For example you write 2 = (3/2) * (4/3) so
log(2)=log(3/2) + log(4/3) = log(1 + 1/2) + log( 1 + 1/3) both of which are easy to calculate. Both take only a few terms in the power series to calculate very accurately.
For 3 you can write 3= 2 * (3/2) so
log(3)=log(2) + log(3/2) and we've already calculated both of these before.
So you can easily calculate log(3)/log(2) which is what OP wants.

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