Express Logs In Terms of t and u

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correction:

ln(x)=t
ln(y)=u

1.
ln(ex/y)-ln(y/(ex)..........use quotient rule

ln(ex)/ln(y)-ln(y)/ln(ex).......use product rule

(ln(e)+ln(x))/ln(y)-ln(y)/(ln(e)+ln(x))............ln(e)=1

(1+ln(x))/ln(y)-ln(y)/(1+ln(x))...........substitute given ln(x)=t and
ln(y)=u

(1+t)/u-u/(1+t)

(t^2 + 2 t - u^2 + 1)/((t + 1) u)


2.
((ln(x))^3-ln(x^4))/((ln(x/e^2))*ln(xe^2))

=(t^3-4ln(x))/((ln(x)-2ln(e))*(ln(x)+2ln(e))).........substitte given ln(x)=t and ln(y)=u

=(t^3-4t)/((t-2*1)*(t+2*1))

=(t(t^2-4))/((t-2)*(t+2))

=(t(t-2)(t+2))/((t-2)*(t+2)).......simplify

=t
 
correction:

ln(x)=t
ln(y)=u

1.
ln(ex/y)-ln(y/(ex)..........use quotient rule

ln(ex)/ln(y)-ln(y)/ln(ex).......use product rule

(ln(e)+ln(x))/ln(y)-ln(y)/(ln(e)+ln(x))............ln(e)=1

(1+ln(x))/ln(y)-ln(y)/(1+ln(x))...........substitute given ln(x)=t and
ln(y)=u

(1+t)/u-u/(1+t)

(t^2 + 2 t - u^2 + 1)/((t + 1) u)


2.
((ln(x))^3-ln(x^4))/((ln(x/e^2))*ln(xe^2))

=(t^3-4ln(x))/((ln(x)-2ln(e))*(ln(x)+2ln(e))).........substitte given ln(x)=t and ln(y)=u

=(t^3-4t)/((t-2*1)*(t+2*1))

=(t(t^2-4))/((t-2)*(t+2))

=(t(t-2)(t+2))/((t-2)*(t+2)).......simplify

=t

I'm surprised to know that I got problem 1 wrong.
Problem 2 is tricky.
 

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