# How to solve the inequality $\log_2(4^x-2(2^x)+17)>5$?

Find the number of positive integers not satisfying the inequality $$\log_2(4^x-2(2^x)+17)>5$$

My approach: let $2^x=t$ then inequality is rewritten in form

$$\log_2(t^2-2t+17)>5$$

then I defined the argument

$$t^2-2t+17>0$$

now I don't know how to proceed next.

Also, I tried to remove log from both sides then I get

$$t^2-2t+17>32$$

$$t^2-2t-15>0$$

solving for $t$ I get $$t=-3,5$$

What to do next?

$$\log _{ 2 }{ (4^{ x }-2(2^{ x })+17)>5 } \\ { 4 }^{ x }-{ 2 }^{ x+1 }+17>32\\ { 2 }^{ 2x }-2{ \cdot 2 }^{ x }-15>0\\ \left( { 2 }^{ x }+3 \right) \left( { 2 }^{ x }-5 \right) >0\\ { 2 }^{ x }\in \left( -\infty ;-3 \right) \cup \left( 5;+\infty \right) \Rightarrow x\in \left( \log _{ 2 }{ 5 } ;+\infty \right)$$ Can you take from here?
• You can create the mathematical symbol for denoting that some element is a member of an interval using \in ( this gives $\in$). Also, do you realize you only gave a partial answer? (Quote from the question: "The question asks to find the number of positive integers not satisfying the inequality [...]"). – wythagoras Jul 17 '16 at 15:10