HELP on Logarithms??? Please!!!

Please be detailed when answering this question.

1. Show that log ( base 2 ) x + log ( base x ) 2-2 can be written as ( a + b )^2 / a.

Note: I think that this question is looking for the values of a and b respectively, which is supposed to be represented with a log value，not the integral value
I've attempted on the method of changing the base of the logarithms. However, I do not know how to proceed.

log ( base 2 ) x + log ( base x ) 2 - 2
(logx)/(log2) +1/[(logx)/(log2)] - 2
a+1/a-2 [(logx)/(log2) = log ( base 2 ) x =a]
(a^2+1-2a)/a
(a^2-2.a.1+1^2)/a
(a^2-2ab+b^2)/a [b=1]
(a-b)^2/a where a = log ( base 2 ) x and b = 1


so it can be written as (a-b)^2/a not ( a + b )^2 / a

log ( base 2 ) x + log ( base x ) 2 -2

Use the change of base rule which implies that log(base y)z=ln z/ln y


ln x/ln 2 + ln2/ln x -2/1 use common denominator which is lnx ln2


[(ln x)^2 +(ln 2)^2 - 2ln x ln2]/ln x ln 2

Let ln x=a and ln 2=b then

[(ln x)^2 - 2ln x ln2 +(ln 2)^2 ]/ln x ln 2

(a^2-2ab+b^2)/ab

=(a^2-b^2)/ab

I don't know how it can be ( a + b )^2 / a