What is the base-10 logarithm of 2?

Question

What is the base-10 logarithm of 2?

2 Answers

Impact of this question

61030 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-28T17:41:29

${log}_{10} 2 = x$ $log_(10)2 = x$
where $10^{x} = 2$ $10^x = 2$

There is no simple way to calculate the value of $x$ $x$
but you can look it up (using a calculator, for example)
${log}_{10} 2 = 0.30103$ $log_(10)2 = 0.30103$ (approx.)

Answer 2 · 2015-05-28T18:19:04

The base-10 logarithm of 2 is the number $x$ $x$ such that $10^{x} = 2$ $10^x = 2$ .

You can calculate logarithms by hand using just multiplication (and dividing by powers of 10 - which is just digit shifting) and the fact that ${log}_{10} (x^{10}) = 10 \cdot {log}_{10} x$ $log_10 (x^10) = 10*log_10 x$ , though it's not very practical...

It goes something like this:

$2 < 10$ $2 < 10$ so ${log}_{10} (2) < 1$ $log_10(2) < 1$

So the part of ${log}_{10} (2)$ $log_10(2)$ before the decimal place is $* 0 *$ $**0**$ .

Calculate $2^{10}$ $2^10$ by multiplying it by itself to get:

$2^{10} = 1024$ $2^10 = 1024$

Notice that $10^{3} = 1000 < 1024 < 10000 = 10^{4}$ $10^3 = 1000 < 1024 < 10000 = 10^4$

So $3 = {log}_{10} (10^{3}) < {log}_{10} (2^{10}) < log (10_{4}) = 4$ $3 = log_10(10^3) < log_10(2^10) < log(10_4) = 4$

Now ${log}_{10} (2^{10}) = 10 \cdot {log}_{10} (2)$ $log_10(2^10) = 10 * log_10(2)$

so by raising to the $10 t h$ $10th$ power, we have multiplied the logarithm by 10.

So the first digit of ${log}_{10} (2)$ $log_10(2)$ after the decimal point is $* 3 *$ $**3**$ .

Then

${log}_{10} (2^{10}) - 3 = {log}_{10} (2^{10}) - {log}_{10} (1000)$ $log_10(2^10) - 3 = log_10(2^10) - log_10(1000)$

$= {log}_{10} (\frac{2^{10}}{1000}) = {log}_{10} (\frac{1024}{1000}) = {log}_{10} (1.024)$ $= log_10(2^10/1000) = log_10(1024/1000) = log_10(1.024)$

So having found the digit $3$ $3$ , we next divide $1024$ $1024$ by $10^{3}$ $10^3$ to get $1.024$ $1.024$ and repeat the process to get the next digit:

${1.024}^{10} ≅ 1.2676506$ $1.024^10 ~= 1.2676506$

This is still less than $10$ $10$ , so the next digit is $* 0 *$ $**0**$ .

${1.2676506}^{10} ≅ 10.715086$ $1.2676506^10 ~= 10.715086$

$10^{1} = 10 \leq 10.715086 < 100 = 10^{2}$ $10^1 = 10 <= 10.715086 < 100 = 10^2$ , so the next digit is $* 1 *$ $**1**$ .

Divide $10.715086$ $10.715086$ by $10^{1}$ $10^1$ to get $1.0715086$ $1.0715086$

${1.0715086}^{10} ≅ 1.99506298$ $1.0715086^10 ~= 1.99506298$

This is less than $10$ $10$ , so the next digit is $* 0 *$ $**0**$ .

${1.99506298}^{10} ≅ 999.0014 ≅ 1000 = 10^{3}$ $1.99506298^10 ~= 999.0014 ~= 1000 = 10^3$

I'll stop here at this approximation, giving a digit $* 3 *$ $**3**$ .

Collecting our digits, we have ${log}_{10} (2) ≅ 0.30103$ $log_10(2) ~= 0.30103$

Science

Math

Humanities

... and beyond

What is the base-10 logarithm of 2?

2 Answers

Related questions

Impact of this question