Question

© 2-(2 raise to 2)2x+1


answer question​

Answer 1

You can proceed through any of the two methods and will get the same answer. The error you made in the second method is not following the cancellation law for inequalities.

(1/2)x2−2x<1/4

loge is an increasing function i.e., x<y⟹logex<logey.  Therefore taking loge both sides is permissible. Had it been a decreasing function, you would not be allowed to do so without altering the inequality.

Taking loge both sides we have

(x2−2x)loge12<loge14

⟹(x2−2x)loge12<2loge12

To get away with the term loge12 from the inequality, you have to divide both sides by the factor loge12. Division is possible if the divisor is non-zero. Here, loge12≠0, the division is permissible. Now, since loge12 is negative, the inequality will be  reversed.

On division we get

(x2−2x)>2.