Question

Solve.
1/2^x > 1/x^2

Answer 1

Clarification on the statement please? Maybe add some parentheses..

Answer 2

\[\frac{ 1 }{ 2^{x} }> \frac{ 1 }{x^{2}}\]

Answer 3

Cross-multiply the proportion then log_base 2 both sides.

Answer 4

how do you log_base 2 both sides?

Answer 5

After cross-multiplying,
\[\large \rightarrow x^2 > 2^x\]
log both sides
\[\large \rightarrow log_2(x^2) > log_2(2^x)\]
simplify
\[\large \rightarrow 2log_2(x) > x\]

Answer 6

the answer is x< -0.77 or 2<x<4
how do yu get that?

Answer 7

This is a pretty complicated situation. A good way to get a handle on it is to graph both functions (either 1/(2^x) and 1/(x^2) or x^2 and 2^x, since both systems will have the same solutions, then you can set the functions equal to each other and solve for the points of intersection).

Answer 8

alright