How do you solve this? #4^x=2^(x-1)-8#

Thanks!

3 Answers
Jun 26, 2018

Answer:

#x notin RR#

Explanation:

#4^x=2^(x-1)-8#
#(2^2)^x=2^(x-1)-8#
#2^(2x)=1/2*2^x-8#
Let #X=2^x#
#X²=1/2X-8#
#X²-1/2X+8#
#Delta=1/4-32#
#Delta=-127/4#
#Delta<0 => x notin RR#
\0/ here's our answer !

Jun 26, 2018

Answer:

No real solutions

Explanation:

#(2^2)^x = 2^(x-1) - 8 #

#=> 2^(2x) - 2^(x-1) + 8 = 0 #

#=> 2^(2x) -(2^(-1) *(2^x)) + 8 = 0 #

#=>( 2^(x))^2 - 1/2 (2^x) + 8 = 0 #

Let # lamda = 2^x #

#=> lamda^2 - 1/2 lamda + 8 = 0 #

#=> " discriminant " < 0 -> "no solutions for " lamda #

Hence no solutions for #lamda # in real, means no solutions for #2^x # hence no solutions for #x #

Jun 26, 2018

Answer:

Continuing:

If you understand complex numbers

Explanation:

#lamda = 1/4 pm sqrt(127)/4i #

#lamda = 2sqrt2 e^(pmiarctan(sqrt(127) ) #

# = 2sqrt2 e^(pmiarctan(sqrt(127) )) e^(2kpii #

as #e^(2kpi i) =1 ,AA k in ZZ #

#=> e^(xln2) = 2sqrt2 e^(i (pmarctan(sqrt(127) )+ 2kpi) #

#=> xln2 = 3/2 ln 2 + i ( 2kpi pm arctansqrt(127) ) #

#=> x = 3/2 + (i ( 2kpi pm arctan sqrt(127) ) )/ln2 #

#AA k in ZZ #