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

Thanks!

3 Answers
Jun 26, 2018

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

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

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