Help with logarithmic problems!

How do I solve the rest of this problem? I'm stuck!

log7(x^2-3x)=log7(4)

and

log2(x-1)=2-log2(3x-2)

1)

log_7(x^2-3x) = log_7(4)

since both logs are to the base 7, you can eliminate logs

=> x^2 - 3x = 4

=> x^2 - 3x - 4 = 0

=> (x - 4)(x + 1) = 0

x = 4 is only valid solution

2)

log_2(x-1) = 2 - log_2(3x-2)

=> log_2(x-1) + log_2(3x-2) = 2

=> log_2 [ (x - 1)(3x - 2) ] = 2

=> 3x^2 - 5x + 2 = 2^2

=> 3x^2 - 5x - 2 = 0

=> 3x^2 - 6x + x - 2 = 0

=> 3x(x - 2) + 1(x - 2) = 0

=> (x - 2)(3x + 1) = 0

only valid solution is x = 2

1) if the base of the logs are identical, you can just set the expressions equal to one another
x^2 - 3x = 4 --> change int a standard for m of quadratic

x^2 - 3x - 4 = 0 -->Is this factorable? yes it is

(x - 4) (x + 1) = 0

x + 1 = 0 --> x = -1 and we cant take the log of a negative number so throw this answer away
x - 4 = 0 --> x = 4

2) log_2 (x - 1) = 2 - log_2 (3x - 2) --> add log_2 (3x -2) to each side

log _ 2 (x - 1) + log_2 (3x - 2) = 2

log_2 ((x - 1)(3x - 2)) = 2

log_2 (3x^2 - 5x + 2) = 2 --> raise everything to the power of 2 --> 2^

2^ (log_2(3x^2 - 5x + 2)) = 2^2 --> log_2 and 2^ are inverse operations and will cancel each other out

(3x^2 - 5x + 2) = 4 --> put this quadratic into standard form

3x^2 - 5x - 2 = 0 ?Is this factorable? yes

(3x + 1)(x - 2) = 0

x - 2 = 0 --> x = 2

3x + 1 = 0 --> x = -1/3. we can never take the log of a negative number and both expressions would be negative for x = 1/3 so toss this answer aside

x = 2 is your answer