Math Problem, can you help me please

Solve this equalization
x € R

(1/2)^x = 4 . 4^x

Thank you

can be written as
2^(-x)=2^(2x+2) { x^a*(x^b)=x^(a+b)}
so we get
-x=2x+2
x=-2/3
hope it helped

(1/2)^x=4.4^x

as 4*4^x=4^x+1

since 4= 2^2 so
4^x+1=2^2(x+1)

so (1/2)^x=2^2(x+1) now take logs at base 2

log(2) (1/2)^x=Log(2) 2^2(x+1)

using log (base a) a^n= n log(base a) a = n and log (a/b) = log a -log b
so
x log (1/2)=2(x+1) log 2
x {log1-log2} = 2x+2
log 1 =0
log 2(base 2) =1
so
-x=2x+2

so x= -2/3

(1/2)^x = 4 . 4^x

Here, we have to represent 1/2 as power of 4.

4^(1/2) = 2
So,
4^(-1/2) = 1/2

But,
(1/2)^x = 4 . 4^x
So,
(4^(-1/2))^x = 4 . 4^x
4^(-x/2) = 4^(1+x)
So,

-x/2=x+1
3x/2 = -1
x=-2/3