|
||||||||||||
|
||||||||||||
| ||||||||||||
Hi Émile, You say "If you had $2^a = 4$ and $2^b = 4$ you could assume that $a = b$ right?" Suppose you didn't want to assume that $a = b$ but rather to prove it. How would you do it? I would start with $2^a = 2^b$ and take the logarithm of both sides to get \[\log\left(2^a\right) = log\left(2^b\right).\] Using a property of the logarithm function this can be rewritten as \[a \log(2) = b \log(2).\] Dividing both sides by $\log(2)$ yields $a = b.$ Now apply the same argument to the equation $1^1 = 1^{0}.$ Taking the logarithm of both sides yields \[\log\left(1^1\right) = log\left(1^0\right)\] or \[1 \log(1) = 0 \log(1).\] But now I am stuck since $\log(1) = 0$ and thus I can't divide both sides by $\log(1).$ All the equation \[1 \log(1) = 0 \log(1)\] says is $0 = 0$ and it doesn't allow me to conclude that $0 = 1.$ I hope this helps, | ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |