Solve 4^(x+1) -16^(x) >2^(3x+2)

Question

Solve the inequality

x < lo g_{2} (22 - 2)

Alternative Form

x \in (- \infty, lo g_{2} (22 - 2))

Evaluate

4^{x + 1} - 1 6^{x} > 2^{3 x + 2}

Move the expression to the left side

4^{x + 1} - 1 6^{x} - 2^{3 x + 2} > 0

Factor the expression

4 (1 - 2^{2 x - 2} - 2^{x}) \times 2^{2 x} > 0

Elimination the left coefficient

(1 - 2^{2 x - 2} - 2^{x}) \times 2^{2 x} > 0

Separate the inequality into 2 possible cases

{1 - 2^{2 x - 2} - 2^{x} > 0 2^{2 x} > 0 {1 - 2^{2 x - 2} - 2^{x} < 0 2^{2 x} < 0

Solve the inequality

More Steps

{x < lo g_{2} (22 - 2) 2^{2 x} > 0 {1 - 2^{2 x - 2} - 2^{x} < 0 2^{2 x} < 0

Since the left-hand side is always positive,and the right-hand side is always 0,the statement is true for any value of x

{x < lo g_{2} (22 - 2) x \in R {1 - 2^{2 x - 2} - 2^{x} < 0 2^{2 x} < 0

Solve the inequality

More Steps

{x < lo g_{2} (22 - 2) x \in R {x > lo g_{2} (22 - 2) 2^{2 x} < 0

Since the left-hand side is always positive,and the right-hand side is always 0,the statement is false for any value of x

{x < lo g_{2} (22 - 2) x \in R {x > lo g_{2} (22 - 2) x \in / R

Find the intersection

x < lo g_{2} (22 - 2) {x > lo g_{2} (22 - 2) x \in / R

Find the intersection

x < lo g_{2} (22 - 2) x \in / R

Solution

x < lo g_{2} (22 - 2)

Alternative Form

x \in (- \infty, lo g_{2} (22 - 2))

Show Solution