Algebra Precalculus – Underlying Reason for Taking Log Base 10

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For the equation $2^x = 7$

The textbook says to use log base ten to solve it like this $\log 2^x = \log 7$.

I then re-arrange it so that it reads $x \log 2 = \log 7$ then divide the RHS by $\log 2$ to isolate the $x$. I understand this part.

I can alternatively solve it in an easier way by simply using $\log_2 7$ on my calculator.

Using both methods the answer comes to the same which is $2.807$

My question is twofold:

  1. Why would the textbook suggest to use log base ten rather than simply using log base two?

  2. I can see how using log base ten and the suggested method in the textbook makes me arrive at the same answer but I don't understand WHY this is so. How does base ten play a factor in the whole scope of things.

Thank you

Best Answer

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

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