Solving an exponential system of equations

[owlsmath]

Great video! I like the very complete explanation of all steps.

[edcagas]

your problems are very interesting and amazing solutions. i love watching your videos

[ScorpioHR]

At 1:45 minute mark it was visible that x=2 and 2y=4 (ergo x=y=2)


3^x*^2y / 5^x * 3^2y = 5^4 * 3^2 / 5^2 * 3^4

Then we can notice that it's

3^x = 3^2
5^2y = 5^4
5^x = 5^2
3^2y = 3^4

And then we see that 2 times x=2 and 2y =4

[Skank_and_Gutterboy]

I did this problem like you did for the first 1:27, then we diverged greatly. At 1:27, you're 99% to a good solution. After transforming the 1st equation, you have (3^x)(5^2y) = (3^2)(5^4). By direct comparison of the exponents (because the bases are the same on the left and right side of the equation), you can deduce that x=2 and 2y=4, therefore x=2 and y=2.

You'll want to make this very same comparison with the 2nd equation. x and y had better be the same or there is something very messed up with this system. In the same manner as above, (5^x)(3^2y) = (5^2)(3^4). So there again, by direct comparison of the exponents, x=2 and 2y=4, so x=2 and y=2. The first and second equations are consistent (if they weren't, the solution would be the "empty set" [also called the "null set"], {x, y} = { }; in other words, no solution).

It's a good idea to check your answer. By direct calculation: (3^2)(25^2) = 5625 and (5^2)(9^2) = 2025, so the solution (x, y) = (2, 2) checks good.