Logarithms: change the base?

Question

if you could at least do one example of each type (A and B), it would be really helpful. Thanks!

solve for x by changing to the appropriate base

A. log(subscript 2)x=log(sub 4)25

log(sub 3)x=log(sub 9)7x-6

log(sub 2)x+log(sub 8)32x=1

solve each equation by changing the base of the logarithm to 10

B. log(sub 2)x=log(sub 5)3

log(sub 7)x=log(sub 2)9

log(sub 2)x+log(sub 4)x=log(sub 2)5

log(sub 3)x=5log(sub 10)2

Answer 1

A.   (log x)/(log 2)  = (log 25)/(log 4)

       log x = (log 25)(log 2)/(log 4)

      log x = (log 25)/2

      log x = log 25^(1/2)

      log x = log 5

         x = 5

    log_3 x = log_9 (7x-6)

    (log x)/(log 3) = [log (7x - 6) ]/(log 9)

    log x = log (7x - 6)/2

   log x = log (7x-6)^(1/2)

     x = (7x-6)^(1/2)

   x^2 = 7x - 6

   x^2 -7x + 6 = 0

  (x-6)(x-1) = 0

   x = 6 or x = 1

i have done 2 from A above.

(log x)/(log 2) = (log 3)/(log 5)

log x = (log 2)(log 3)/(log 5)