Hello,
Please show what you've tried
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