show that if , then
I tried to solve this with change of base but didn't seem to help. so not sure even what the first step is on this one.
After all replacements and multiplying with log_{a}(c-b) and log_{a}(c+b) you get this:
log_{a}(c-b)+log_{a}(c+b) = 2
Then you can apply the log law giving:
log_{a}((c-b)(c+b)) = 2
which simplifies to:
log_{a}(c^2-b^2) = 2
Now use the definition of logs to get this:
a^2 = c^2 - b^2 Which is the same as c^2 = a^2 + b^2
It would take a better mind than mine to prove so without doing the work the other way. But my comment was that the problem statement said to start with a^2 + b^2 = c^2 and derive the log statement. So the answer should be in that form. So saying, when I did the problem I did it 'backward" like most people would, but then I reversed it to get the "correct" derivation. Nothing I could see tells me that going from the log statement to a^2 + b^2 = c^2 implied that the converse was true until I actually worked the steps.
-Dan