I posted a thread about proving the following: [itex]b^(log_b x) = x[/itex] a while back, and recieved help in understanding multiple ways to prove it. But I also asked a math teacher prior to requesting help here, he suggested 2 methods: Substitution(also suggested by my physics teacher): Let [itex]log_b x = y[/itex], which is the same as writing [itex]b^y = x[/itex], right? but we already said y = [itex]log_b x = y[/itex], therefore, by substituting we get: [itex]b^(log_b x) = x[/itex] That method was simple enough, I was also shown by the people here that by applying that [itex]log_b x[/itex] and [itex]b^x[/itex] were inverses, you could achieve the same final result: let [itex]log_b x = f(x)[/itex], and [itex]b^x = g(x)[/itex], by composing the inverses you get: g(f(x)) = x = [itex]b^f^(x)[/itex] = [itex]b^(log_b x)[/itex] And after much work, I finally understood this method as well. But another way the math teacher suggested, was what he called the 'intuitive' way, simply by 'looking' at the problem, you could see the relationship to be true... my guess is that he meant you could see the first way mentioned here simply by 'looking'. Does anyone have an idea of what he may have been suggesting? Thanks in advance.