Underlying Reason For Taking Log Base 10

Solution 1:

This works in any base because $\log_b (a^c) = c \log_b(a)$

The practical reason for using base $10$ was a little old fashioned: it allowed the use of tables of logarithms instead of a calculator and reducing these calculations to addition and subtraction. Calculators then provided a $\log_{10}$ function as a carry-over from tables and just called the button log

My 1953 elementary statistical tables have logarithms of numbers from $1$ to $10$ and anti-logarithms of numbers from $0$ to $1$. Since these are logarithms base $10$, I can easily deal with all numbers because $\log(a \times 10^n)=n +\log(a)$

Here I want $\dfrac{\log 7}{\log 2}$.

• My tables tell me $\log 7\approx 0.8451_6$ (with the ${\,}_6$ helping interpolation) and $\log 2\approx 0.3010_{22}$. So that leaves me with trying to calculate $\dfrac{0.8451}{0.3010}$. I cannot be bothered to do long division, so instead I try to calculate $\text{antilog}\,({\log 0.8451 -\log 0.3010})$.

• My tables tell me $\log 8.45 \approx 0.9269_5$ so I write $\log 0.8451 \approx \bar{1}.9269$ (with the $\bar{1}$ because I wanted $\log (8.451\times 10^{-1})= -1+\log 8.451$). Similarly it tells me $\log 3.01 \approx 0.4786_{14}$ so I write $\log 0.3010 \approx \bar{1}.4786$

• I now calculate by hand $\bar{1}.9269 - \bar{1}.4786 = 0.4483$. My tables tell me $\text{antilog}\,0.448 \approx 2.805_7$ and I interpolate the final $3$ using the ${\,}_7$ to give a final approximate answer of $2.807$. Which is what you got with some clever silicon

Solution 2:

Question 1) As you see, both method arrive at the same answer, however $\log$ base 10 is a natural choice, as stated in the comment by Matti P above.

Question 2) It is because for $a>0,a\neq 1,b>0, c>0, c\neq 1$,$$\log_a b=\frac{\log_c b}{\log_c a}$$In particular in your example, $c=10$, so both method arrives the same answer. The formula is known as the change of base formula, see here.

Solution 3:

Logarithmic functions enjoy many properties.

One of the very interesting properties of logarithms is the formula called change of base formula.

$$ \log_a x = \frac {\log_b x}{\log_b a} $$

For example $$ \log_2 7 = \frac {\log_{10} 7}{\log_{10} 2} $$

This formula makes finding logarithms in an arbitrary base possible by using only logarithms base $10$ or natural logarithms.

Solution 4:

The use of Log in base 10 is a common choice and we often find it in calculators since our standard system of counting is in base 10 (i.e. number of fingers).

To handle algebraic problems any other choice is fine depending upon the specific problem to solve by logarithmic identities.

For the application in calculus the best choice is the natural logarithm in base e.

Solution 5:

Other answers mention the change of base $$ \log_a x = \frac {\log_b x}{\log_b a}. $$ An immediate consequence is that a quotient of logarithms does not depend on the base: $$\frac {\log_ax}{\log_ay}=\frac {\frac {\log_bx}{\log_ba}}{\frac {\log_by}{\log_ba}} =\frac {\log_bx}{\log_by}. $$