Basically, I don't get exponential equations. Could someone please help me? These are the two I'm having the most trouble with:
2^(1 - 2x) = 3x
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.