# log identity

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• May 4th 2011, 01:53 PM
bigwave
log identity
show that if $\displaystyle {a}^{2}+{b}^{2}={c}^{2}$, then

$\displaystyle {log}_{b+c}a + {log}_{c-b}a= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)$

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, 01:58 PM
Moo
Hello,

Please show what you've tried :)
• May 4th 2011, 02:19 PM
bigwave
fyi i made a correction on the org post with the bases
• May 4th 2011, 02:29 PM
poirot
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, 02:31 PM
bigwave
using change of base
Quote:

Originally Posted by bigwave
show that if $\displaystyle {a}^{2}+{b}^{2}={c}^{2}$, then

$\displaystyle {log}_{b+c}a + {log}_{c-b}a= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)$

ok so change of base would be

$\displaystyle \frac{log(a)}{{log}(b+c)}} + \frac{log(a)}{{log}(c-b)}}= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)$

but how can you add these?
• May 4th 2011, 02:39 PM
VincentP
Try changing all Logs to log_{a}. And remember log_{a}a=1.
• May 4th 2011, 02:44 PM
bigwave
Quote:

Originally Posted by poirot
try {log}_{b+c}a^2+{log}_{c-b}b^2.

hopefully you can work out what that's meant to be.

$\displaystyle {log}_{b+c}a^2+{log}_{c-b}b^2.$

or

$\displaystyle 2{log}_{b+c}a+2{log}_{c-b}b$

not sure where we are going with this.
• May 4th 2011, 02:51 PM
bigwave
Quote:

Originally Posted by VincentP
Try changing all Logs to log_{a}. And remember log_{a}a=1.

what do you do with the coeficient 2
• May 4th 2011, 02:52 PM
poirot
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, 03:05 PM
VincentP
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, 03:17 PM
topsquark
Technically, you know, this should be the other way around. The question says that if $\displaystyle a^2 + b^2 = c^2$, so you need to also be sure you know how to reverse the argument.

-Dan
• May 4th 2011, 03:23 PM
poirot
Thats what my method did topsquark
• May 4th 2011, 03:31 PM
topsquark
Quote:

Originally Posted by poirot
Thats what my method did topsquark

Sorry, I forgot to quote. My comment was regarding VincentP's post. And, I think bigwave didn't get that point either.

-Dan
• May 4th 2011, 03:34 PM
VincentP
Doesn't the proof just show that there is a biconditional relation between the statements?
• May 4th 2011, 04:15 PM
topsquark
Quote:

Originally Posted by VincentP
Doesn't the proof just show that there is a biconditional relation between the statements?

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
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