Question 664573
clever professor - mixes an exponential problem with a simultaneous equation problem.  you have my sympathy


the key to the solution is knowing that x^a / x^b = x^(a-b)
once you know that you're a little less than half way there.


the problem statement is to solve the following equations simnultaneously.


2^x - 3^y = -1
2^x-2 - 3^y-1 = -1 


2^x-2 is equivalent to 2^x / 2^2 which is equivalent to 2^x / 4


3^y-1 is equivalent to 3^y/3^1 which is equivalent to 3^y / 3


substitute in your original equations and they become:


2^x - 3^y = - 1
2^x/4 - 3^y/3 = - 1


we will solve this simultaneous by substitution.


solve for 2^x in the first equation to get 2^x = 3^y - 1


substitute for 2^x in the second equation to get ((3^y) - 1)/4 - 3^y/3 = - 1


multiply both sides of the equation by 12 to get:


12 * ((3^y) - 1) / 4 - 12 * (3^y)/3 = 12 * - 1


simplify to get:


3 * ((3^y)-1) - 4 * (3^y) = - 12


simplify further to get:


3 * (3^y) - 3 - 4 * (3^y) = - 12


combine like terms to get:


- (3^y) - 3 = - 12


add 3 to both sides of the equation to get:


- (3^y) = - 9


multiply both sides of this equation by -1 to get 3^y = 9


since 3^2 = 9, y must be equal to 2 **********************


in the equation 2^x = 3^y - 1, replace y with 2.
this makes the equation 2^x = 3^2 - 1 which makes the equation 2^x = 9 - 1 which makes the equation 2^x = 8


since 2^3 = 8, then x must be equal to 3 **********************


you have x = 3 and y = 2 and all that remains is to go back to the original equation and confirm that these solutions are good.


the original equations are:


2^x - 3^y = -1
2^x-2 - 3^y-1 = -1 


replace  with 3 and y with 2 in these equations to get:


2^3 - 3^2 = -1
2^3-2 - 3^2-1 = -1 


simplify these equations to get:


2^3 - 3^2 = -1 becomes 8 - 9 = -1 which becomes -1 = -1 which is true.
2^3-2 - 3^2-1 = -1 becomes 2^1 - 3^1 = -1 which becomes 2 - 3 = -1 which becomes -1 = -1 which is also true.


the values for x and y solve both original equations so they're good.