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 ). #