A)

2 log^2(x)-5 log(x)-3 = 0

Substitute u = log(x):

2 u^2-5 u-3 = 0

Solve the quadratic equation by completing the square:

Divide both sides by 2:

u^2-(5 u)/2-3/2 = 0

Add 3/2 to both sides:

u^2-(5 u)/2 = 3/2

Add 25/16 to both sides:

u^2-(5 u)/2+25/16 = 49/16

Factor the left hand side:

(u-5/4)^2 = 49/16

Take the square root of both sides:

abs(u-5/4) = 7/4

Eliminate the absolute value:

u-5/4 = -7/4 or u-5/4 = 7/4

Add 5/4 to both sides:

u = -1/2 or u-5/4 = 7/4

Substitute back for u = log(x):

log(x) = -1/2 or u-5/4 = 7/4

Cancel logarithms by taking exp of both sides:

x = 1/sqrt(e) or u-5/4 = 7/4

Add 5/4 to both sides:

x = 1/sqrt(e) or u = 3

Substitute back for u = log(x):

x = 1/sqrt(e) or log(x) = 3

Cancel logarithms by taking exp of both sides:

x = 1/sqrt(e) or x = e^3

Now test that these solutions are appropriate by substitution into the original equation:

Check the solution x = 1/sqrt(e):

2 log^2(x)-5 log(x)-3 => -3-5 log(1/sqrt(e))+2 log^2(1/sqrt(e)) = 0

So the solution is correct.

Check the solution x = e^3:

2 log^2(x)-5 log(x)-3 => -3-5 log(e^3)+2 log^2(e^3) = 0

So the solution is correct.

Thus, the solutions are:

x = 1/sqrt(e) or x = e^3

B)

(log(x))/(log(2)) = (log(18))/(log(2))-(log(6))/(log(2))

Divide both sides by 1/(log(2)):

log(x) = log(2) ((log(18))/(log(2))-(log(6))/(log(2)))

Cancel logarithms by taking exp of both sides:

x = 3

Now test that this solution is appropriate by substitution into the original equation:

Check the solution x = 3:

(log(x))/(log(2)) => (log(3))/(log(2)) = (log(3))/(log(2)) ~~ 1.58496

(log(18))/(log(2))-(log(6))/(log(2)) => -(log(6))/(log(2))+(log(18))/(log(2)) = (log(3))/(log(2)) ~~ 1.58496

So the solution is correct.

Thus, the solution is:

x = 3

Basically, if you see a quadratic pattern like with A), e.g. in the form ax^2+x+c, where c=some constant, then solve using quadratic equation or completing square. Best to make sure you have your log rules down too:

Mathwords: Logarithm Rules

Hope that helps? Best of luck! ("problem 2's" picture appears to not be working...not displaying correctly?)