Question

Given log 102 = 0.3010 and log 103= 0.4771, find log 1049 without using a calculator.

Answer 1

Bearing in mind that when the base is omitted, base 10 is assumed, or namely log(x) = log₁₀(x).

`\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( (x)/(y)\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\log_(10)(2)=0.3010\qquad \qquad \log_(10)(3)=0.4771 \\\\[-0.35em] ~\dotfill\\\\ \log_(10)\left( \cfrac{4}{9} \right)\implies \log\left( \cfrac{4}{9} \right)\implies \log\left[ \cfrac{2^2}{3^2} \right]\implies \log\left[ \left( \cfrac{2}{3} \right)^2 \right]\implies 2\log\left[ \cfrac{2}{3}\right] \\\\\\ 2[\log(2)-\log(3)]\implies 2[0.3010~~ - ~~0.4771]\implies -0.3522`