How do you solve this? #4^x=2^(x-1)-8#
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"What balanced equation represents a redox reaction?"
#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 !
#(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 #
Continuing:
If you understand complex numbers
#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 #
