Hi ! I did it so (2^x)-(3^x)=((2^(2x))*((2^x)-(3^x))^(1/3).Now lets take t=(2^x)-(3^x).Now take to the power 3 and you get t^3=((2)^2x))*t.Now 0 is obvious solution.Now divide by t ,you get t^2=2^2x.Now evaluate with x and you get (3/2)^x=2 and at the end x=log2/(log3-log2).It is less stresa ful method 😉.Greetings from Slovenia.
Hi ! I did it so (2^x)-(3^x)=((2^(2x))*((2^x)-(3^x))^(1/3).Now lets take t=(2^x)-(3^x).Now take to the power 3 and you get t^3=((2)^2x))*t.Now 0 is obvious solution.Now divide by t ,you get t^2=2^2x.Now evaluate with x and you get (3/2)^x=2 and at the end x=log2/(log3-log2).It is less stresa ful method 😉.Greetings from Slovenia.