What is the value of log 43?

#43# is a prime number, so #log 43# cannot be expressed in terms of logarithms of smaller numbers.

Using a calculator, we find:

#log 43 ~~ 1.63346845558#

How could we find it by hand?

One somewhat arduous method goes as follow:

Note that #10^1 <= 43 < 10^2#, so the portion of the logarithm before the decimal point must be #color(red)(1)#

Dividing #43# by #10# (and thus subtracting #1#) from the logarithm, we get #4.3#.

If we raise this to the tenth power, then its logarithm will be multiplied by #10#, i.e. shifted one place to the left.

We find:

#4.3^10 = 2161148.2313284249#

So:

#10^6 <= 4.3^10 < 10^7#

So the next digit of the logarithm is #color(red)(6)#

Divide #2161148.2313284249# by #10^6# to subtract #6# from its logarithm and raise to the tenth power to shift the logarithm another place to the left:

#2.1611482313284249^10 ~~ 2222.519#

Then:

#10^3 <= 2222.519 < 10^4#

so the next digit is #color(red)(3)#

Keep on going for as many digits as you want.

Thus far we have found:

#log 43 ~~ 1.63#

Suppose you know that:

#log 2 ~~ 0.30103#

#log 3 ~~ 0.47712#

Then note that:

#43 = 129/3 ~~ 128/3 = 2^7/3#

So

#log 43 ~~ log(2^7/3) = 7 log 2 - log 3 ~~ 7*0.30103-0.47712 = 1.63009#

We know that the error is approximately:

#log (129/128) = log 1.0078125 = (ln 1.0078125) / (ln 10) ~~ 0.0078/2.3 = 0.0034#

So we can confidently give the approximation:

#log 43 ~~ 1.633#

A calculator tells me:

#log 43 ~~ 1.63346845558#