# Logarithm Examples

Here you will learn some logarithm examples for better understanding of logarithm concepts.

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

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

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.