Question

If ${\log _a}bc = x$, ${\log _b}ac = y$ and ${\log _c}ab = z$, then $\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}} = ?$
A. 0
B. 1
C. $\dfrac{1}{2}$
D. None

Answer 1

Hint- When it comes to questions having algorithms, there are a few properties which we use. Out of those properties, we will use two. Those two are ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ and $\log \left( {a.b} \right) = \log a + \log b$. Using these two properties, we will solve our question.

Complete Step-by-Step solution:
The given values of x, y, and z in the question are-
$
  x = {\log _a}bc \\
    \\
  y = {\log _b}ac \\
    \\
  z = {\log _c}ab \\
$
Applying the property ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ in all three equations mentioned above we get,
$
  x = \dfrac{{\log bc}}{a} \\
    \\
  y = \dfrac{{\log ac}}{b} \\
    \\
  z = \dfrac{{\log ab}}{c} \\
$
After applying this property, we will simplify it to further extent and apply another property $\log \left( {a.b} \right) = \log a + \log b$ which will get us-
$
  x = \dfrac{{\log b + \log c}}{{\log a}} \\
    \\
  y = \dfrac{{\log c + \log a}}{{\log b}} \\
    \\
  z = \dfrac{{\log a + \log b}}{{\log c}} \\
$
So we get the simplified values of x, y and z.
The given equation in the question is $\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$. We will simply put the values of x, y, and z in this given equation and solve it further.
We will solve the given equation in three parts in order to make is easy and simple.
Part 1-
$\dfrac{1}{{x + 1}}$
Putting the value of x, we get-
$\dfrac{1}{{\dfrac{{\log b + \log c}}{{\log a}} + 1}}$
Taking LCM and solving it further, we get-
$\dfrac{{\log a}}{{\log b + \log c + \log a}}$ $ \to $ equation 1
Part 2-
$\dfrac{1}{{y + 1}}$
Putting the value of y, we get-
$\dfrac{1}{{\dfrac{{\log c + \log a}}{{\log b}} + 1}}$
Taking LCM and solving it further, we get-
$\dfrac{{\log b}}{{\log c + \log a + \log b}}$ $ \to $ equation 2
Part 3-
$\dfrac{1}{{z + 1}}$
Putting the value of z, we get-
$\dfrac{1}{{\dfrac{{\log a + \log b}}{{\log c}} + 1}}$
Taking LCM and solving it further, we get-
$\dfrac{{\log c}}{{\log a + \log b + \log c}}$ $ \to $ equation 3
Now, we will add the equation1, equation 2, and equation 3, we will have-
$
  \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}} \\
    \\
   \Rightarrow \dfrac{{\log a + \log b + \log c}}{{\log a + \log b + \log c}} \\
    \\
   \Rightarrow 1 \\
$
Hence, the value of $\dfrac{1}{{x + 1}} + \dfrac{1}{{y + 1}} + \dfrac{1}{{z + 1}}$ is equal to 1.
Thus, option number B is correct.
Note: Whenever you solve algorithm based questions, keep in mind to apply the properties in order to make it solvable. Without the properties, it will be very confusing to solve and will become very hectic for you only.