To solve the second equation you can write:
(You can also take the logarithm of both side in the beginning and continue)
Now use (the general rule)
About the first equation, are you sure it isn't in stead of ?
Do you understand that the logarithm is the inverse function to the exponential? And, it turns out you can use logarithm to any base!
You also need to learn the "rules of expentials":
a^x a^y= a^(x+ y) and (a^x)^y= a^(xy)
and the corresponding "rules of exponentials:
log(ab)= log(a)+ log(b) and log(a^b)= b log(a).
It is that last one that is of the most use.
Taking the logarithm of both sides of that second equation, as Siron suggests, gives
log((7/9)^x)= x log(7/9)= (1-x)log(11). You now have a linear equation to solve for x.
I agree with Siron also that you should look again at the first problem. Although that can be solve, I doubt it is what is intended.
2^(1- 2x)= 2(2^(-2x))= 2(1/4)^x= 3x. dividing both sides by 3 and Multiplying both sides by (1/4)^x you get x4^x= 2/3. Of course, 4= e^ln(4) so 4^x= e^(xln(4)) and the equation is xe^(xln(4))= 2/3. Now multiply both sides by ln(4) so the equation is (xln(4))e^(xln(4))= 2ln(4)/3. Almost there! Let y= x ln(4). The equation ye^y= 2ln(4)/3 has solution y= W(2ln(4)/3) where "W(x)" is the "Lambert W function"- which is defined as the inverse function to f(x)= xe^x. Since y= W(2ln(4)/3) and y= x ln(4), x= W(2ln4)/3)/ln(4).
But that is NOT the sort of problem anyone who is not in a "special functions" course where they are introduced to the W function would be expected to do!
If the problem is in fact 2^(2x-1)= 3^x, take the logarithm of both sides:
(1- 2x)log(2)= x log(3). Again, you have a linear equation to solve for x.