Thnx guys... i really appreciate your help

However, if you move one term to the other side of the equal sign, then you can get a step-by-step solution.

Solve for x:
9^(3 x-2) = 7^(6 x-4)
Look for an exponent to divide both sides by.
Divide both sides by 9^(3 x-2):
1 = 7^(6 x-4) 9^(2-3 x)
Reverse the equality in 1 = 7^(6 x-4) 9^(2-3 x) in order to isolate x to the left hand side.
1 = 7^(6 x-4) 9^(2-3 x) is equivalent to 7^(6 x-4) 9^(2-3 x) = 1:
7^(6 x-4) 9^(2-3 x) = 1
Write exponents in 7^(6 x-4) 9^(2-3 x) = 1 in terms of a common base.
7^(6 x-4) 9^(2-3 x) = e^(log(7^(6 x-4))) e^(log(9^(2-3 x))) = e^((6 x-4) log(7)) e^((2-3 x) log(9)) = exp((6 x-4) log(7)+(2-3 x) log(9)):
exp(log(7) (6 x-4)+log(9) (2-3 x)) = 1
Eliminate the exponential from the left hand side.
Take the natural logarithm of both sides:
log(7) (6 x-4)+log(9) (2-3 x) = (2 i) pi n  for  n element Z
Write the linear polynomial on the left hand side in standard form.
Expand and collect in terms of x:
(6 log(7)-3 log(9)) x-4 log(7)+2 log(9) = (2 i) pi n  for  n element Z
Isolate terms with x to the left hand side.
Subtract 2 log(9)-4 log(7) from both sides:
(6 log(7)-3 log(9)) x = 4 log(7)-2 log(9)+(2 i) pi n  for  n element Z
Solve for x.
Divide both sides by 6 log(7)-3 log(9):
Answer: | 
| x = (4 log(7))/(6 log(7)-3 log(9))-(2 log(9))/(6 log(7)-3 log(9))+((2 i) pi n)/(6 log(7)-3 log(9))  for  n element Z

In fact it is 9^y==7^(2 y) which of course you see is (3^2)^y == 7^(2 y) which is 3^(2 y)==7^(2 y) which is 3^z ==7^z you can do by hand only ...