Solve this?

#|2^(2x)-9|> 5#

2 Answers
Feb 27, 2018

# x = 1 or log_2 sqrt14#

Explanation:

#|2^(2x)-9|=5#

So , #+-[2^(2x)-9]=5#

Take

#+(2^(2x)-9)=5#

#=>2^(2x)=14#

#=>log_2 14=2x#

#=>1/2log_2 14=x#

#=>log_2 14^(1/2)=x#

#=>log_2 sqrt14=x#

Or

#-(2^(2x)-9)=5#

#=>-2^(2x)+9=5#

#=>4=2^(2x)#

#=>2^2=2^(2x)#

#=>2=2x#

#=>1=x#

Feb 27, 2018

#1 >x >1.90# approximately

#1>x>ln(14)/(2ln(2))# exactly

Explanation:

#color(blue)("Considering initial condition")#

The absolute type brackets are still just brackets. Special ones admittedly. They state the any calculation that occurs inside is to be interpreted as ending up as positive

Example #|-2|=|+2|=+2#

So if we set #y=|2^(2x)-9|# we can only accept the value of #x# that give #y>5#

So if the content of the absolute is set at #b# we can only accept

#|color(white)(./.)+5 < b < -5color(white)(./.)|#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the critical point 1")#

Set the critical point 1 as #P_1->y_("critical")=-5=2^(2x)-9#

#-5+9=2^(2x)#

#4=2^(2x)#

We know that #2^2=4# so

#2^(2x) =2^2# thus #x=1#

As #x# becomes less then #2^(2x)# becomes less. Also #2^(2x)-9# becomes more negative. So we have #x<1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the critical point 2")#
Set the critical point 2 as #P_2->y_("critical")=+5=2^(2x)-9#

#14<2^(2x)#

#ln(14)<2xln(2)#

#x> ln(14)/(2ln(2))#

#x~~1.903677.......#
#x~~1.90# to 2 decimal places #larr" Reasonable answer"#

#2^(2xx1.90) ~~13.928...#
#2^(2xx1.91)~~14.123...#

As #x# gets greater than #1.91" then "{2^(2xxx)-9}# becomes greater than +5

So we have #x>1.90# to 2 decimal places
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

#1 >x >1.90# approximately

#1>x>ln(14)/(2ln(2))# exactly

#color(magenta)("Graph showing how it all ties together")#
Tony B