# Math Help - log identity

1. ## log identity

show that if ${a}^{2}+{b}^{2}={c}^{2}$, then

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

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.

2. Hello,

Please show what you've tried

3. fyi i made a correction on the org post with the bases

4. try {log}_{b+c}a^2+{log}_{c-b}b^2.

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

5. ## using change of base

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

${log}_{b+c}a + {log}_{c-b}a= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)$
ok so change of base would be

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

but how can you add these?

6. Try changing all Logs to log_{a}. And remember log_{a}a=1.

7. 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.
${log}_{b+c}a^2+{log}_{c-b}b^2.$

or

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

not sure where we are going with this.

8. 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

9. 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)

10. 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

11. Technically, you know, this should be the other way around. The question says that if $a^2 + b^2 = c^2$, so you need to also be sure you know how to reverse the argument.

-Dan

12. Thats what my method did topsquark

13. 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

14. Doesn't the proof just show that there is a biconditional relation between the statements?

15. 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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