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
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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(5315)
Your working start is good:
+
+
= +
+
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:
==
=
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.
Your working start is good: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: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.
You can put this solution on YOUR website!
