3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

$$\log _{ 2 }{ (4^{ x }-2(2^{ x })+17)>5 } \\ { 4 }^{ x }-{ 2 }^{ x+1 }+17>32\\ { 2 }^{ 2x }-2{ \cdot 2 }^{ x }-15>0\\ \left( { 2 }^{ x }+3 \right) \left( { 2 }^{ x }-5 \right) >0\\ { 2 }^{ x }\in \left( -\infty ;-3 \right) \cup \left( 5;+\infty \right) \Rightarrow x\in \left( \log _{ 2 }{ 5 } ;+\infty \right) $$ Can you take from here?

$\endgroup$
2
  • 1
    $\begingroup$ 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 [...]"). $\endgroup$ – wythagoras Jul 17 '16 at 15:10
  • $\begingroup$ @wythagoras,thanks for editing,I know,it is not a whole answer $\endgroup$ – haqnatural Jul 17 '16 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.