# How to use substitution to solve for x? ln^2 x-11lnx+30=0

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The equation (ln x)^2-11*lnx+30=0 has to be solved for x.

This is a quadratic equation with the variable ln x.

(ln x)^2-11*lnx+30=0

(ln x)^2- 5*lnx - 6*ln x+30=0

(ln x)(ln x - 5) -6(ln x - 5)=0

(ln x - 6)(ln x - 5) = 0

ln x= 6 and ln x = 5

For ln x = 6, x = e^6 and for ln x = 5, x = e^5

The solution of the given equation are x = e^6 and x = e^5

We'll substitute ln x by another variable.

ln x = t

We'll re-write the equation in t:

t^2 - 11t + 30 = 0

We'll solve the quadratic:

t1 = [11+sqrt(11^2 - 4*30)]/2

t1 = [11+sqrt(121 - 120)]/2

t1 = (11 + 1)/2

t1 = 12/2

t1 = 6

t2 = (11 - 1)/2

t2 = 5

Now, we'll solve for x:

ln x = t1 => ln x = 6 => x = e^6

ln x = t2 => ln x = 5 => x = e^5

**The values of x that represent the solutions of the equation are: {e^5 ; e^6}.**