Solve for x : \(2^{x + 2}\) > \(({1\over 4})^{1\over x}\).

Solution :

We have \(2^{x + 2}\) > \(2^{2/x}\).

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

(the sign of 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  \(\implies\)  x \(\in\) \((0, \infty)\)


Similar Questions

Find the value of \(2log{2\over 5}\) + \(3log{25\over 8}\) – \(log{625\over 128}\).

Evaluate the given log : \(81^{l\over {log_5 3}}\) + \(27^{log_9 36}\) + \(3^{4\over {log_7 9}}\).

If \(log_a x\) = p and \(log_b {x^2}\) = q then \(log_x \sqrt{ab}\) is equal to

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 \(({x\over y})^{b-c}\) \(\times\) \(({y\over z})^{c-a}\) \(\times\) \(({z\over x})^{a-b}\).

Leave a Comment

Your email address will not be published. Required fields are marked *

Ezoicreport this ad