Logarithm Examples

Example 1 : If $log_e x$ – $log_e y$ = a, $log_e y$ – $log_e z$ = b & $log_e z$ – $log_e x$ = c, then find the value of $({x\over y})^{b-c}$ $\times$ $({y\over z})^{c-a}$ $\times$ $({z\over x})^{a-b}$.

Solution : $log_e x$ – $log_e y$ = a $\implies$ $log_e {x\over y}$ = a $\implies$ $x\over y$ = $e^a$

$log_e y$ – $log_e z$ = b $\implies$ $log_e {y\over z}$ = b $\implies$ $y\over z$ = $e^b$

$log_e z$ – $log_e x$ = c $\implies$ $log_e {z\over x}$ = c $\implies$ $z\over x$ = $e^c$

$\therefore$     $(e^a)^{b-c}$ $\times$ $(e^b)^{c-a}$ $\times$ $(e^c)^{a-b}$

=     $e^{a(b-c)+b(c-a)+c(a-b)}$ = $e^0$ = 1

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Example 2 : If $log_a x$ = p and $log_b {x^2}$ = q then $log_x \sqrt{ab}$ is equal to-

Solution : $log_a x$ = p $\implies$ $a^p$ = x $\implies$ a = $x^{1/p}$

Similarly     $b^q$ = $x^2$ $\implies$ b = $x^{2/q}$

Now,     $log_x \sqrt{ab}$ = $log_x \sqrt{x^{1/p}x^{2/q}}$ = $log_x x^{({1\over p}+{2\over q}){1\over 2}}$ = $1\over {2p}$ + $1\over q$.

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Example 3 : Solve for x : $2^{x+2}$ > $({1\over 4})^{1/x}$

Solution : We have $2^{x+2}$ > $2^{-2/x}$.

Since the base 2 > 1, we have x + 2 > $-2\over x$

(the sign of inequality is retained)

Now     x + 2 + $-2\over x$ > 0 $\implies$ ${x^2 + 2x + 2}\over x$ > 0

$\implies$     $({x+1})^2 + 1\over x$ > 0

$\implies$     x $\in$ (0,$\infty$).

Practice these given logarithm examples to test your knowledge on concepts of logarithm.