Question

If x, y, z are in H.P then prove that
\[\log \left( x+z \right)+\log \left( x+z-2y \right)=2\log \left( x-z \right)\]

Answer 1

Hint: In this question, we first need to get the relation between the x, y, z from the given H.P condition which is given by \[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y}\]. Then from the left hand side of the given equation we can apply the formula of \[\log a+\log b=\log ab\] and then multiply the two terms inside the logarithm and write them as one logarithm term. Now, on applying the factorisation of polynomials formula after expanding the terms we get the result.

Complete step-by-step solution -
HARMONIC PROGRESSION (H.P):
A sequence \[{{a}_{1}},{{a}_{2}},....,{{a}_{n}}\]of non-zero numbers is called a Harmonic Progression (H.P), if the sequence \[\dfrac{1}{{{a}_{1}}},\dfrac{1}{{{a}_{2}}},.....,\dfrac{1}{{{a}_{n}}}\]is an A.P
Now, if the terms a, b, c are in H.P then we can express them as
\[\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b}\]
Here, given that the terms x, y, z are in H.P
Now, we get the relation between them from the above formula as
\[\Rightarrow \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y}\]
Now, this can be further written as
\[\Rightarrow \dfrac{x+z}{xz}=\dfrac{2}{y}\]
Now, on cross multiplication we get,
\[\Rightarrow \left( x+z \right)y=2xz\]
Now, from the given equation in the question we have
\[\log \left( x+z \right)+\log \left( x+z-2y \right)=2\log \left( x-z \right)\]
Now, to prove the above equation let us consider the left hand side
\[\Rightarrow \log \left( x+z \right)+\log \left( x+z-2y \right)\]
As we already know from the properties of logarithm that
\[\log a+\log b=\log ab\]
Now, using the properties of logarithm we can write the above expression as
\[\Rightarrow \log \left( \left( x+z \right)\left( x+z-2y \right) \right)\]
Now, on expanding the terms further inside the logarithm we get,
\[\Rightarrow \log \left( {{\left( x+z \right)}^{2}}-2y\left( x+z \right) \right)\]
As we already know from the given H.P condition on x, y, z we get,
\[\Rightarrow \left( x+z \right)y=2xz\]
Now, on substituting this in the above expression we get,
\[\Rightarrow \log \left( {{\left( x+z \right)}^{2}}-2\times 2xz \right)\]
Now, on simplifying it further we get,
\[\Rightarrow \log \left( {{\left( x+z \right)}^{2}}-4xz \right)\]
As we already know from the factorisation of polynomials that
\[{{\left( a+b \right)}^{2}}-4ab={{\left( a-b \right)}^{2}}\]
Now, on substituting the above formula in the expression we get,
\[\Rightarrow \log \left( {{\left( x-z \right)}^{2}} \right)\]
As we already know from the properties of logarithm that
\[\log {{a}^{b}}=b\log a\]
Now, using this formula the above expression can be further written as
\[\Rightarrow 2\log \left( x-z \right)\]
Hence, proved that \[\log \left( x+z \right)+\log \left( x+z-2y \right)=2\log \left( x-z \right)\].

Note: Instead of considering the left hand side we can also solve it by considering the left hand side and then converting accordingly using the properties of logarithm and factorisation into the known relation between x, y, z.
It is important to note that instead of considering the relation between x, y, z using H.P as \[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y}\] if we consider it as \[x+z=2y\] then the corresponding expression changes and so the final result. Then the result obtained will be completely incorrect.