lg^2(2x)-9lg(x)+2=0 pleasee ?

I'll interpret it slightly differently:

(log 2x)^2 - 9.log x + 2 = 0

[(log 2) + (log x)]^2 - 9.log x + 2 = 0

Expanding that ^2 term,

(log 2)^2 + 2.log 2.log x + (log x)^2 - 9.log x + 2 = 0

Collecting terms,

(log x)^2 + (2.log 2 - 9).log x + (log 2)^2 + 2 = 0

Solving this quadratic in log x

log x = {-(2.log 2 - 9) ± √{(2.log 2 - 9)^2 - 4((log 2)^2 + 2)}} ÷ 2

log x = {-(2.log 2 - 9) ± √{4(log 2)^2 - 36.log 2 + 81 - 4(log 2)^2 - 8}} / 2

Reminder, log is log base e,

this gives us two values for log x, from which we can find x.

The values for x are roughly 1441 and roughly 1.5

I did work them out exactly and verified they solved the original equation, but I leave

the precise values up to the student to work out again as a calculator exercise. 

log x = -log 2 + 9/2  ± √{-9.log 2 + 73/4}

lg^2(2x)-9lg(x)+2=0

I assume that this is...

ln (2x)^2 - 9 ln x + 2 = 0  .... so we have

2ln(2x) - 9 ln (x) + 2 = 0

2ln(2) + 2ln(x) - 9ln(x) + 2 = 0     factor

-7(ln x) = -2 - 2 ln(2)  =

7 ln(x) = 2 + 2ln(2)

ln (x)  = [2 + 2ln(2)] / 7

In exponential form, we have

e ^ ((2 + 2ln(2))/7) = x = about 1.622

Addendum: that final line in the analysis has got displaced, it belongs just above the line

Reminder, log is log base e ...⏰