Question

If $\ln (x + z) + \ln (x - 2y + z) = 2\ln (x - z)$ , then
A.$y = \dfrac{{2xz}}{{x + z}}$
B.${y^2} = xz$
C.$2y = x + z$
D.$\dfrac{{x + y}}{{y - z}} = \dfrac{x}{z}$

Answer 1

Hint: Note that there is no logarithm in the options. So start by simplifying the expression in a manner that will let you take the antilogarithm. After doing that, proceed for further simplification of the resulting equation to arrive at any one of the given options.

Complete step-by-step answer:
Recall the property of logarithm which states $a\ln b = \ln {b^a}$. Using this property we can express the above equation as $\ln (x + z) + \ln (x - 2y + z) = \ln {(x - z)^2}$
Now we will use the property of addition of logarithms which states $\ln a + \ln b = \ln ab$. Using this, we will get,
$\ln (x + z)(x - 2y + z) = \ln {(x - z)^2}$
We will now take antilog on both sides. That will give us,
$(x + z)(x - 2y + z) = {(x - z)^2}$
We just need to simplify this now. We do this by multiplying the first two terms and then using the formula ${(a - b)^2} = {a^2} + {b^2} - 2ab$ for the term on the right hand side.
$
   \Rightarrow {{{x}}^2} + - 2xy + xz + zx - 2zy + {{{z}}^2} = {{{x}}^2} + {{{z}}^2} - 2xz \\
   \Rightarrow 2xz - 2xy - 2zy = - 2xz \\
   \Rightarrow 4xz = 2xy + 2zy \\
   \Rightarrow 2xz = xy + zy \\
   \Rightarrow 2xz = y(x + z) \\
   \Rightarrow y = \dfrac{{2xz}}{{x + z}} \\
 $
So, as we can see, the correct answer is option A.

Note: Since there is no logarithm term in any of the options, start the question by simplifying the expression such that there is only one logarithmic term on each side. This is necessary for taking the anti-logarithm. After taking the anti-logarithm, you will get a pure algebraic equation. This just needs to be simplified. Once it is simplified, you can check the final answer with the options provided and determine the correct option.