Tag: introduction to logarithm

Given, $\log _c a=x$

(a) Since $log _a a = 1,$ so $log _{10}  10 = 1$
(b) $log (2 + 3) log 5$ and $log (2 \times 3) = log 6 = log 2 + log 3$
$\therefore log(2 + 3) \neq log (2 \times 3)$
(c) Since, $log _a  1 = 0$, so, $log _{10} 1 = 0$.
(d) $log(1 + 2 + 3) = log 6 = log (1 \times 2 \times 3) = log 1 + log 2 + log 3$

$log _{10} 80 = log _{10} (8 \times 10)$
$= log _{10} 8 + log _{10} 10$
$=log _{10} (2^3) + 1$
$= 3 log _{10} 2 + 1$
$= (3 \times 0.3010) + 1$
$= 1.9030$

$log _5 512 = \dfrac{log 512}{log 5}$
$=\dfrac{log 2^9}{log (10/2)}$
$=\dfrac{9 log  2}{log 10 - log 2}$
$=\dfrac{9 \times 0.3010}{1- 0.3010}$
$= \dfrac{2.709}{0.699}$
$=\dfrac{2709}{699}$
$= 3.876$

$log (2^{64}) = 64 \times log 2$
$= (64 \times 0.30103)$
$= 19.2592$
Its characteristic is 19.
Hence, then number of digits in $2^{64}$ is 20.

$log _x \left( \dfrac{9}{16} \right ) = - \dfrac{1}{2}$
$\Rightarrow x^{-1/2} = \dfrac{9}{16}$
$\Rightarrow \dfrac{1}{\sqrt x} = \dfrac{9}{16}$
$\Rightarrow \sqrt x = \dfrac{16}{9}$
$\Rightarrow x = \left( \dfrac{16}{9} \right)^2$
$\Rightarrow x = \dfrac{256}{81}$

The value of $\dfrac {1}{2}\log _{10} 25 - 2 \log _{10} 3 +\log _{10} 18$ is
$= \log _{10}(25)^{1/2} - \log _{10} (3)^{2} + \log _{10}18$
$= \log _{10}5 - \log _{10}9 + \log _{10}18$
$= \log _{10} \left (\dfrac {5}{9}\times 18\right ) $

Let $log _2      16 = n$
Then, $2^n = 16 = 2^4    \Rightarrow n = 4$
$\therefore log _2  16 = 4$

$5^2=25$

Exponential form of $\log _2 16=4$