How to calculate $10^{0.4}$ without using calculator

Question

How to calculate $10^{0.4}$ without using calculator or if not what is the closest answer you can get just using pen and paper within say $2$ min?

Answer 1

When doing rough logarithms or antilogarithms under conditions like the ones in the question, it helps to remember a few frequently-used approximations: \begin{align} 10^{0.301} & \approx 2 \\ 10^{0.5} & \approx 3.16 \end{align} It may be easier to remember the first fact if you keep in mind that $2^{10} = 1024$ is a little greater than $10^3$.

Then observe that $$10^{0.4} = \frac{10}{\left(10^{0.3}\right)^2} \approx \frac{10}{2^2} = 2.5. \tag1$$

If you want to refine this, note that the rough approximation $10^{0.3} \approx 2$ has an error between $2\%$ and $2.5\%$, so Equation $(1)$ has divided by too much and the result should be about $4\%$ or $5\%$ greater than shown.

Answer 2

$x = 10^{0.4} = 10^{2/5} = \sqrt[5]{100}$

In other words, you want to solve $x^5 = 100$.

Wolfram alpha says that $$x \approx 2.511886431509580111085032067799327394158518100782475428679...$$

Let $f(x) = x^5$

Then $f(x + h) \approx f(x) + h f'(x)$ becomes $(x+h)^5 \approx x^5 + 5hx^4$.

So, if $x$ is an approximate answer to $x^5 = 100$, then we could get a better approximation, $x+h$ by solving

\begin{align} (x + h)^5 &= 100 \\ x^5 + 5hx^4 &= 100 \\ 5hx^4 &= 100 - x^5\\ h &= \dfrac{100 - x^5}{5x^4} \end{align}

So, if $x_n$ is your most recent approximation to the solution to $x^5 = 100$, then your next approximation would be \begin{align} x_{n+1} &= x_n + \dfrac{100 - x_n^5}{5x_n^4} \\ &= x_n \left(1 + \dfrac{100 - x_n^5}{5x_n^5}\right) \\ &= x_n \left(1 + \dfrac{20}{x_n^5} - \dfrac 15\right) \\ &= x_n \left(0.8 + \dfrac{20}{x_n^5}\right) \\ \end{align}

We can start with a linear approximation first

\begin{array}{c|c} x & x^5\\ \hline 2 & 32 \\ 2+h & 100 \\ 3 & 243\\ \hline \end{array}

where we get \begin{align} \dfrac{(2+h) - 2}{3-2} &= \dfrac{100 - 32}{243 - 32}\\ h &= \dfrac{68}{211}\\ h &\approx 2.3 \end{align}

so we start with $x_1 = 2.3$

As a general BOE(back of the envelope) computation rule, you double the number of digits after the decimal point in each approximation. Since $2.3$ has one digit after the decimal point, we will round off to the nearest humdredth for our first approximation.

\begin{align} 2.3^2 &= 5.29 \\ 2.3^4 &= 5.29^2 \\ 2.3^4 &= 27.98 \\ 2.3^5 &= 2.3 \cdot 27.98 \\ 2.3^5 &= 64.35 \\ 20/2.3^5 &= 0.31\\ x_2 &= 2.3(0.8 + 0.31)\\ x_2 &= 2.55\\ \end{align}

Next, we will round off to four digits past the decimal point.

\begin{align} 2.55^2 &= 6.5025 \\ 2.55^4 &= 6.5025^2 \\ 2.55^4 &= 42.2825 \\ 2.55^5 &= 2.55 \cdot 42.2825 \\ 2.55^5 &= 107.8204 \\ 20/2.55^5 &= 0.1855\\ x_2 &= 2.55(0.8 + 0.1855)\\ x_2 &= 2.5130\\ \end{align}

The next step, if we had performed it, would give

2.51188742

which is accurate to 6 digits.

Answer 3

You are looking for a quick method that requires the minimum of arithmetic. In the absence of fifth powers of integers which are close to 100, we look for two integers the ratio of the fifth powers of which are close to 100, and we are fortunate that $$\frac{5^5}{2^5}=\frac{3125}{32}=\frac{3200}{32}-\frac{75}{32}$$

We can therefore consider a simple linear approximation of the form $$f(x+h)\simeq f(x)+hf'(x)$$ using the function $$f(x)=x^{\frac 15}\Rightarrow f'(x)=\frac 15 x^{-\frac 45}$$

Therefore $$100^{\frac 15}=f(100)=f\left(\frac{3125}{32}+\frac{75}{32}\right)\simeq 2.5+\frac {75}{32}\times\frac 15\times\left(\frac{3125}{32}\right)^{-\frac 45}$$ $$\simeq 2.5+\frac{15}{32}\times\frac{16}{625}$$ $$\simeq 2.5+\frac{3}{250}$$ $$\simeq 2.512$$

This is correct to 3 decimal places and easily doable in two minutes.