# How do I solve 'log(base 10) 5' without using the calculator?

Apr 16, 2016

See explanation

#### Explanation:

If you have memorized that
$\log 2 = 0.3$
you can follow this way
$\log 5 = \log \left(\frac{10}{2}\right) = 1 - \log 2 = 1 - 0.3 = 0.7$

If you want a general way to find logarithms without using calculators or tables, you could use this formula:
$\left(\frac{1}{2}\right) \ln | \frac{1 + x}{1 - x} | = f \left(x\right) = x + {x}^{3} / 3 + {x}^{5} / 5 + \ldots$
And
$\log y = \ln \frac{y}{\ln} 10 = \frac{2}{\ln} 10 \cdot \left(\frac{1}{2} \cdot \ln | y |\right)$ => $\log y = 0.869 \cdot \left(\frac{1}{2} \cdot \ln | y |\right)$ where $y = \frac{1 + x}{1 - x}$
(Note1: you can use $\frac{2}{\ln} 10 = 0.868589$ with the precision you like. Using two terms of the series, 0.869 has a proper level of precision. Note 2: the values of x must be smaller than 1.)

We can't calculate $\log 5$ directly because
$\frac{x + 1}{1 - x} = 5$ => $x + 1 = 5 - 5 x$ => $6 x = 4$ => $x = 1.5$
And the series doesn't converge when $x > 1$

But since $5 = 2 \cdot 2.5$
for ${y}_{1} = 2 \to \frac{x + 1}{1 - x} = 2$ => $x + 1 = 2 - 2 x$ => $x = \frac{1}{3} \cong 0.3333$
$f \left(x = \frac{1}{3}\right) = \frac{1}{3} + \frac{1}{3} ^ 3 \cdot \frac{1}{3} = \frac{1}{3} + \frac{1}{81} = 0.3333 + 0.0123 = 0.3456$

for ${y}_{2} = 2.5 \to \frac{x + 1}{1 - x} = 2.5$ => $x + 1 = 2.5 - 2.5 x$ => $3.5 x = 1.5$ => $x = \frac{3}{7} \cong 0.4286$
Of course we can use this $x = 0.4286$. But perhaps there is an easier way (without a calculator we need to think of this) such as:

Considering that $5 = {2}^{2} \cdot 1.25$ (and since we have already calculated $f \left(x = \frac{1}{3}\right)$):
for ${y}_{2} = 1.25 \to \frac{x + 1}{1 - x} = 1.25$ => $x + 1 = 1.25 - 1.25 x$ => $2.25 x = 0.25$ => $x = \frac{25}{225} = \frac{1}{9} \cong 0.1111$
$f \left(x = \frac{1}{9}\right) = 0.1111 + \frac{1}{9} ^ 3 \cdot \frac{1}{3} = 0.1111 + \frac{1}{729} \cdot \frac{1}{3} = \frac{1}{9} + \frac{1}{2187} = 0.1111 + 0.0005 = 0.1116$
(as to the number $0.0005$ just remember that $\frac{10}{2} = 5$)

Using the results above
$\log 5 = 0.869 \left[2 \cdot \left(\frac{1}{2} \cdot \ln | 2 |\right) + \left(\frac{1}{2} \cdot \ln | 1.25 |\right)\right] = 0.869 \left[2 \cdot f \left(x = \frac{1}{3}\right) + f \left(x = \frac{1}{9}\right)\right] = 0.869 \left[2 \cdot 0.3456 + 0.1116\right] = 0.869 \left[0.6912 + 0.1116\right] = 0.869 \cdot 0.8028 = 0.6976332$ or $0.698$ in 3 decimals

We should be aware that this last estimate is smaller than the correct result.

(In fact $\log 5 = 0.6990$)