Solve for b if: 4log base2 2x + 2log base b x = 4

Log base b=logb

4 log₂ 2x+ 2logb  x=4
log₂(2x)⁴+logb x²=log₂16
log₂ 16x⁴+ logb x²=log₂ 16
logb x²=log₂ 16 - log₂ 16x⁴
logb x²=log₂ (16/16x⁴)
logb x²= log₂ x⁻⁴

logb x²=log₂ x² / log₂ b

log₂ x² / log₂ b=-log x⁴
log₂ b=log₂ x² /- log₂ x⁴
log₂ b=2 log x / -4 log₂ x
log₂ b=-1/2    ⇔ b=2⁻¹/² =1/√2=(√2)/2≈  0.707106781

b=(√2)/2     

To chek

if x=10
log base √2/2= log√2/2
4 log₂ 20 + 2log√2/2  10=4(4.321928095...)+2 (-6.64385619=
=  17.28771238-13.28771238=4

I’m not sure how this works