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.

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- May 4th 2011, 02:53 PMbigwavelog identity
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. - May 4th 2011, 02:58 PMMoo
Hello,

Please show what you've tried :) - May 4th 2011, 03:19 PMbigwave
fyi i made a correction on the org post with the bases

- May 4th 2011, 03:29 PMpoirot
try {log}_{b+c}a^2+{log}_{c-b}b^2.

hopefully you can work out what that's meant to be. - May 4th 2011, 03:31 PMbigwaveusing change of base
- May 4th 2011, 03:39 PMVincentP
Try changing all Logs to log_{a}. And remember log_{a}a=1.

- May 4th 2011, 03:44 PMbigwave
- May 4th 2011, 03:51 PMbigwave
- May 4th 2011, 03:52 PMpoirot
sorry that b^2 should be another a^2 so 2(log(a)/log(b+c)+log(a)/log(c-b))and add the fractions. Then its a simple case of log laws. log(x)+log(y)=log(xy)

- May 4th 2011, 04:05 PMVincentP
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 - May 4th 2011, 04:17 PMtopsquark
Technically, you know, this should be the other way around. The question says that if , so you need to also be sure you know how to reverse the argument.

-Dan - May 4th 2011, 04:23 PMpoirot
Thats what my method did topsquark

- May 4th 2011, 04:31 PMtopsquark
- May 4th 2011, 04:34 PMVincentP
Doesn't the proof just show that there is a biconditional relation between the statements?

- May 4th 2011, 05:15 PMtopsquark
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