# How do I find the number whose common logarithm is 2.6025?

Jul 14, 2015

Can use ${\log}_{10} \left(2\right) \cong 0.30103$ to find approximate value $400$ then multiply by an approximation for ${10}^{0.00044} \cong 1.001$ to get $400.4$.

If you have a calculator then just ${10}^{2.6025} \cong 400.4055$

#### Explanation:

${10}^{2.6025} = {10}^{2} \cdot {10}^{0.60206} \cdot {10}^{0.00044}$

$\cong 100 \cdot {10}^{2 {\log}_{10} \left(2\right)} \cdot {10}^{0.00044}$

$= 100 \cdot {2}^{2} \cdot {10}^{0.00044}$

$= 400 \cdot {10}^{0.00044}$

Now

${1.001}^{1000} = {\left(1 + 0.001\right)}^{1000}$

$= 1 + \left(\begin{matrix}1000 \\ 1\end{matrix}\right) 0.001 + \left(\begin{matrix}1000 \\ 2\end{matrix}\right) {0.001}^{2} + \ldots$

$= 1 + 1 + 0.4995 + 0.166167 + 0.04141712475 + \ldots$

somewhere between $2.7$ and $3$

${\log}_{10} \left(3\right) \cong 0.4771$ (one of those useful numbers to memorise)

${\log}_{10} \left(2.7\right) = {\log}_{10} \left(\frac{27}{10}\right) = {\log}_{10} \left({3}^{3} / 10\right) = 3 {\log}_{10} \left(3\right) - 1$

$\cong 0.4313$

So $0.4313 < 1000 {\log}_{10} \left(1.001\right) < 0.4771$

$0.0004313 < {\log}_{10} \left(1.001\right) < 0.0004771$

So $1.001$ is a fairly good approximation for ${10}^{0.00044}$