SOLUTION: Show that [1/loga (abc)] + [1/logb (abc)] + [1/logc (abc)] = 1 My working: [1/log abc/log a] + [1/log abc/log b] + [1/log abc/log c] =[log a/log abc] + [log b/log abc] + [lo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Show that [1/loga (abc)] + [1/logb (abc)] + [1/logc (abc)] = 1 My working: [1/log abc/log a] + [1/log abc/log b] + [1/log abc/log c] =[log a/log abc] + [log b/log abc] + [lo      Log On


   

Question 771674: Show that [1/loga (abc)] + [1/logb (abc)] + [1/logc (abc)] = 1
My working:
[1/log abc/log a] + [1/log abc/log b] + [1/log abc/log c]
=[log a/log abc] + [log b/log abc] + [log c/log abc]
=[log (a*b*c)] / [log (abc)^3]
=log (abc) / log (abc)^3
=1/3

My answer is is 1/3, it's wrong. I should prove that the answer is equals to 1. Please help me.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1%2Flog%28a%2C%28abc%29%29+1%2Flog%28b%2C%28abc%29%29+1%2Flog%28c%2C%28abc%29%29=1

Your working start is good:
1%2F%28%28log%28abc%29%2Flog%28a%29%29%29 + 1%2F%28%28log%28abc%29%2Flog%28b%29%29%29 + 1%2F%28%28log%28abc%29%2Flog%28c%29%29%29
=log%28%28a%29%29%2Flog%28%28abc%29%29 + log%28%28b%29%29%2Flog%28%28abc%29%29 + log%28%28c%29%29%2Flog%28%28abc%29%29
Then, you forget that you are adding fractions with a common denominator, and multiply the denominators instead of leaving the common denominator alone, and just adding the numerators,like this:
=%28log%28%28a%29%29%2Blog%28%28b%29%29%2Blog%28%28c%29%29%29%2Flog%28%28abc%29%29=log%28%28abc%29%29%2Flog%28%28abc%29%29=1
That kind of confusion happens to most of us now and then.
Unfortunately, once we make a mistake like that we cannot detect it by ourselves.
We need help from a pair of fresh, unbiased eyes.