Solve the inequation a^(2x)−a^(2)a^(x)+a^(x)−a^(2) < 0 for all a ∈ RR_+ without {1} ?

1 Answer
Mar 2, 2018

\qquad \qquad \qquad \qquad \qquad "solution set" \ = ( -oo, 2 ).

Explanation:

"We want to solve the inequality:"

\qquad \qquad \qquad \qquad \qquad \qquad a^{ 2 x } - a^2 a^x + a^x - a^2 < 0; \qquad \qquad \ \ \ a \in RR^{+} - \{ 0 \}.

\qquad \qquad \qquad \qquad \qquad \ ( a^{ x } )^2 - a^2 a^x + a^x - a^2 < 0;

"Notice -- the expression on the left can be factored !!!"

\qquad \qquad \qquad \qquad \qquad \qquad \quad ( a^x - a^2 ) (a^x + 1 ) < 0;

"The quantity" \ a^x \ "is always positive, as" \ a \ "is given positive, and is"
"used as the base of an the exponential expression:"

\qquad \qquad \qquad \qquad \qquad \qquad \ ( a^x - a^2 ) \underbrace{ (a^x + 1 ) }_{ "always postive" } < 0;

"The product of the two factors on the left-hand side of the"
"above inequality is negative. The right factor is always"
"positive. Thus, the left factor must be always negative."

\qquad :. \qquad \qquad \qquad \qquad \qquad \ a^x - a^2 < 0;

\qquad :. \qquad \qquad \qquad \qquad \qquad \qquad a^x < a^2;

\qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \quad x < 2.

"So the solution set of the given inequality, in interval notation,"
"is:"

\qquad \qquad \qquad \qquad \qquad \qquad \quad "solution set" \ = ( -oo, 2 ).