|
Obviously, T>U, so write T=U+W. We have then 2^U+W-3^U=1, or 2^W.(10/5)^U-3^U=1, so 2^W.10^U=15^U+5^U. Now, for U>2, if U is odd the right hand term ends in 500 (because 15^U ends in 375 and 5^U ends in 125) -- two zeroes, and if U is even, both 15^U and 5^U end in 25 -- one zero. On the other hand, the left hand term ends in U zeroes. So, the only possible cases that manage the same number of zeroes are U=0 and U=1, which are the given solutions. |
