Logarithm Examples

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



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,\(\infty\)).


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