Logarithm Questions

Solution : \(2log{2\over 5}\) + \(3log{25\over 8}\) + \(log{128\over 625}\) = \(log{2^2\over 5^2}\) + \(log({5^2\over 2^3})^3\) + \(log{2^7\over 5^4}\) = \(log({2^2\over 5^2}{5^6\over 2^9}{2^7\over 5^4})\) = log 1 = 0 Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the given log : \(81^{l\over {log_5 3}}\) + \(27^{log_9 36}\) + \(3^{4\over […]

Solution : We have \(2^{x + 2}\) > \(2^{2/x}\). Since the base 2 > 1, we have x + 2 > -\(2\over x\) (the sign of the inequality is retained) Now x + 2 + \(2\over x\)  \(\implies\) \(x^2 + 2x + 2\over x\) > 0 \(\implies\) \((x + 1)^2 + 1\over x\) > 0 

Solution : \(81^{log_3 5}\) + \(3^{3log_9 36}\) + \(3^{4log_9 7}\) \(\implies\) \(3^{4log_3 5}\) + \(3^{log_3 {(36)}^{3/2}}\) + \(3^{log_3 {7}^2}\) = 625 + 216 + 49 = 890. Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\). If \(log_a x\)

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, \(log_x \sqrt{ab}\) = \(log_x \sqrt{x^{1/p}x^{2/q}}\) = \(log_x x^{({1\over p}+{2\over q}){1\over 2}}\) = \(1\over {2p}\) + \(1\over q\). Similar Questions Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\). Evaluate the

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 \(\implies\) \(z\over x\) =