@@HarshPatel-jl2cnNice solution! Tho we could ask if this is the only solution, its clear that if x would be greater than 5, both 2^(x-3) and 3^(2x-8) would be greater than than 36, and the same for smaller x value. A more "standard" method would be to make the exponential expression have the same exponent. 3^(2x-8)=3^(2(x-4))=9^x-4. From here multiplying both side by 9, we get [2^(x-3)]×[9^(x-4)]=324. From here the base will be 18, and solving by using logarithm, you get the foretold solution of 5.
I got this question in my olympiad exam and i didn't know how to solve it : [2^(x-3)] × [3^(2x-8) ]=36 ; find x
36 can be written as 2^2x3^2 and therefore 2^x-3 = 2^2 and 3^2x-8 = 3^2. Therefore x-3=2 and 2x-8=2 and solving both equations we get x=5
@@HarshPatel-jl2cnNice solution! Tho we could ask if this is the only solution, its clear that if x would be greater than 5, both 2^(x-3) and 3^(2x-8) would be greater than than 36, and the same for smaller x value. A more "standard" method would be to make the exponential expression have the same exponent. 3^(2x-8)=3^(2(x-4))=9^x-4. From here multiplying both side by 9, we get [2^(x-3)]×[9^(x-4)]=324. From here the base will be 18, and solving by using logarithm, you get the foretold solution of 5.