Question

If \[x = {\log _a}bc,y = {\log _b}ca,z = {\log _c}ab\] then
A.\[\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}} = 1\]
B.\[\dfrac{1}{{x - 1}} + \dfrac{1}{{y - 1}} + \dfrac{1}{{z - 1}} = 1\]
C.\[xyz = x + y + z + 1\]
D.\[xyz = 1\]

Answer 1

Hint: Given are some values which on simplification and on certain operations will result in a new equation. We need to find from the option which one is correct. For that we will first simplify the values given in the form of x, y and z using the laws of logarithms. Then we will check from the options which one is suitable.

Complete step-by-step answer:
Given that,
\[x = {\log _a}bc,y = {\log _b}ca,z = {\log _c}ab\]
We know that, \[{\log _n}m = \dfrac{{\log m}}{{\log n}}\] and \[\log mn = \log m + \log n\] then the combination of these two rules is used above. Using them we will separate the logs as follows,
\[x = \dfrac{{\log b + \log c}}{{\log a}},y = \dfrac{{\log c + \log a}}{{\log b}},z = \dfrac{{\log a + \log b}}{{\log c}}\]
This is the simplified ratio we can say.
Now, if we observe that each of them is having log with a, b and c but situated in a different way.
Option C and D will be simply eliminated because both sides will never be equal.
We will first try for option A.
\[\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}\]
Now putting the values,
\[ = \dfrac{1}{{\dfrac{{\log b + \log c}}{{\log a}} + 1}} + \dfrac{1}{{\dfrac{{\log c + \log a}}{{\log b}} + 1}} + \dfrac{1}{{\dfrac{{\log a + \log b}}{{\log c}} + 1}}\]
We need to take the LCM here in the denominator,
\[ = \dfrac{1}{{\dfrac{{\log b + \log c + \log a}}{{\log a}}}} + \dfrac{1}{{\dfrac{{\log c + \log a + \log b}}{{\log b}}}} + \dfrac{1}{{\dfrac{{\log a + \log b + \log c}}{{\log c}}}}\]
Taking the denominator of the denominator in the numerator we get,
\[ = \dfrac{{\log a}}{{\log b + \log c + \log a}} + \dfrac{{\log b}}{{\log c + \log a + \log b}} + \dfrac{{\log c}}{{\log a + \log b + \log c}}\]
If we observe now the denominators of all the terms are the same, just the terms are in different positions. So we will arrange them as,
\[ = \dfrac{{\log a}}{{\log a + \log b + \log c}} + \dfrac{{\log b}}{{\log a + \log b + \log c}} + \dfrac{{\log c}}{{\log a + \log b + \log c}}\]
Now since the denominator is same we can directly add the numerator terms,
\[ = \dfrac{{\log a + \log b + \log c}}{{\log a + \log b + \log c}}\]
Since the numerator and denominator are same we will cancel them,
\[ = 1\]
Thus option A is correct.
So, the correct answer is “Option A”.

Note: Note that when we cancel the numerator and denominator that are the same that won’t mean the answer is zero. It is zero if the same terms are subtracted.
Also note that why option B is not correct because the minus in the denominator would make one of the terms different. And the answer will not be equal to 1.