prove that log(a/b) + log(b/a)=log(a+b)?

Question

I want to prove that log (a/b) + log (b/a) = log (a+b) using any way.

Answer 1

I agree with jcherry, ... but here's something interesting to consider.

Let's find the values of a and b that make the expression true!



log(a/b) + log(b/a) = log(a+b)

log [ a/b • b/a ] = log (a+b)

log [1] = log (a+b)

0 = log (a+b)

a + b = 1



The statement log(a/b) + log(b/a)=log(a+b)

is true when a + b = 1 where a and b are positive.



I hope this helps!

Answer 2

log (a/b) + log (b/a) = log a - log b + log b - log a = 0.



False. Let a = b = 1. Then log (a/b) + log (b/a) = log 1 + log 1 = 0 + 0 = 0 while log (a+b) = log 2 and clearly, log 2 =/= 0.

Answer 3

log(a/b) = log(a) - log(b)

log(b/a) = log(b) - log(a)



log(a) - log(b) + log(b) - log(a)

= 0



no....that's a no go right there.



Put more simply, the log of any number will equal the additive inverse of its own reciprocal. So that expression is definitely wrong. Unless a + b = 1, but since that hasn't been explicitly defined in the problem, then it's invalid to assume that it does.

Answer 4

You want to prove something that can never truly be proven.



log(a/b) + log(b/a) = log(a) – log(b) + log(b) – log(a) = 0.



Unless a+b = 1, log(a+b) does not equal 0. Let's just use a = 5 and b = 3. Log(8) does not equal 0.

Answer 5

log(a/b) + log(b/a)=log(a+b)

log[(a/b)*(b/a)] = log(a+b)

[(a/b)*(b/a)] = (a+b)

a + b = 1

It's true for each couple of (a,b both > 0) when a + b = 1

ex:

a = 1/3

b = 2/3

log((1/3)/(2/3)) + log((2/3)/(1/3)) = log1

log(1/2) + log(2) = log(1)

Answer 6

log a - log b + log b - log a =0 don't see how.

Answer 7

log(a/b) + log(b/a)=log(a+b) ??

Are you sure you wrote the equation correctly?



log(a/b) + log(b/a)

=log (a/b)(b/a)

= log 1

= 0



which is not log(a+b).

Answer 8

loga/b + log b/a

= loga - log b + log b - loga =0 & not log(a+b)