Question

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

Thanks!

Answer 1

`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 !

Answer 2

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`

Answer 3

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`