Question

If $x$ , $y$ , $z$ are in HP, then $\log \left( {x + z} \right) + \log \left( {x - 2y + z} \right)$ is equal to
$1){\text{ }}\log \left( {x - z} \right)$
$2){\text{ 2}}\log \left( {x - z} \right)$
$3){\text{ 3}}\log \left( {x - z} \right)$
$4){\text{ 4}}\log \left( {x - z} \right)$

Answer 1

Hint : If three terms a, b, c are in HP (Harmonic Progression), then $\dfrac{1}{a}$ , $\dfrac{1}{b}$ and $\dfrac{1}{c}$ form an A.P. Therefore, the harmonic mean formula is $\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$ and the harmonic mean is $b = \dfrac{{2ac}}{{a + c}}$ . By using this we will get the value of y then use this value of y in the given expression. By looking at the options given in the question we came to know that there is only one log term in all the options but in the given equation of question there are two log terms.

Complete step-by-step answer :
It is given that x, y, z are in harmonic progression (HP) which means that $\dfrac{1}{x}$ , $\dfrac{1}{y}$ , $\dfrac{1}{z}$ are in A.P .
So we have to convert two log terms into one and this can be done by using the properties of logarithms. After doing this we have to make the equation simpler. And then we have to come back to the properties of the logarithms that an exponent with a variable can be multiplied in the front. After applying all these steps we get the desired answer.
By using harmonic mean formula we have $\dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}$
By taking L.C.M at the right hand side of the above equation we get , $\dfrac{2}{y} = \dfrac{{x + z}}{{xz}}$ which gives us harmonic mean as $y = \dfrac{{2xz}}{{x + z}}$ --------(i)
The given expression in the question is $\log \left( {x + z} \right) + \log \left( {x - 2y + z} \right)$ . So
$\log \left( {x + z} \right) + \log \left( {x - 2y + z} \right)$
By applying the property , $\log m + \log n = \log \left( {mn} \right)$ we get
$\Rightarrow \log \left[ {\left( {x + z} \right)\left( {x - 2y + z} \right)} \right]$
Put the value of y given in the equation (i) ,
$\Rightarrow {\text{ }}\log \left[ {\left( {x + z} \right)\left( {x - 2\left( {\dfrac{{2xz}}{{x + z}}} \right) + z} \right)} \right]$
$\Rightarrow {\text{ }}\log \left[ {\left( {x + z} \right)\left( {x - \dfrac{{4xz}}{{x + z}} + z} \right)} \right]$
$\Rightarrow {\text{ }}\log \left[ {\left( {x + z} \right)\left( {\dfrac{{x\left( {x + z} \right) - 4xz + \left( {x + z} \right)z}}{{x + z}}} \right)} \right]$
The terms $\left( {x + z} \right)$ will be cancelled out ,
$\Rightarrow {\text{ }}\log \left[ {x\left( {x + z} \right) - 4xz + \left( {x + z} \right)z} \right]$
$\Rightarrow {\text{ }}\log \left( {{x^2} + xz - 4xz + xz + {z^2}} \right)$
Adding the similar terms, we ge
$\Rightarrow {\text{ }}\log \left( {{x^2} - 2xz + {z^2}} \right)$
As we know that ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$ . Therefore,
$\Rightarrow {\text{ }}\log {\left( {x - z} \right)^2}$
By using property , $\log {m^n} = n\log m$ , we get
$\Rightarrow {\text{ 2}}\log \left( {x - z} \right)$
Hence, the correct option is $2){\text{ 2}}\log \left( {x - z} \right)$
So, the correct answer is “Option 2”.

Note : By reading the question we came to know that this question deals with the harmonic progressions and logarithmic so it’s clear that we have to keep in mind the properties of harmonic progression, arithmetic progressions because harmonic progressions are linked with arithmetic progressions and the properties of logarithms. In order to solve a problem on harmonic progression, one should make the corresponding AP series and then solve the problem. Three consecutive numbers of a harmonic progression are $\dfrac{1}{{a - d}}$ , $\dfrac{1}{a}$ , $\dfrac{1}{{a + d}}$ .