13.if X^(2)+y^(2)=23xy Prove That Log(x+y)/(5)=(1)/(2)(log X+log Y)

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Jun 28, 2024

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** XIII. Proof of Logarithmic Identity **

In this article, we will prove the following logarithmic identity:

$\frac{\log (x+y)}{5} = \frac{1}{2} (\log x + \log y)$

given the condition:

$x^2 + y^2 = 23xy$

Step 1: Rearrange the given equation

Let's start by rearranging the given equation:

$x^2 + y^2 = 23xy$

We can rewrite this as:

$x^2 - 23xy + y^2 = 0$

Step 2: Factorize the equation

Now, let's factorize the equation:

$x^2 - 23xy + y^2 = (x - y)^2 - 22xy = 0$

This implies that:

$(x - y)^2 = 22xy$

Step 3: Take the logarithm

Take the logarithm of both sides:

$\log (x - y)^2 = \log (22xy)$

Using the property of logarithms, we can rewrite this as:

$2\log (x - y) = \log 22 + \log x + \log y$

Step 4: Simplify the equation

Now, let's simplify the equation:

$2\log (x - y) = \log 22 + \log x + \log y$

Subtract $\log x + \log y$ from both sides:

$2\log (x - y) - \log x - \log y = \log 22$

Step 5: Use logarithmic properties

Using the quotient rule of logarithms, we can rewrite the left-hand side as:

$\log \frac{(x-y)^2}{xy} = \log 22$

Step 6: Simplify further

Simplify the equation:

$\log (x+y)^2 = \log 22 + \log xy$

Divide both sides by 2:

$\log (x+y) = \frac{1}{2} \log 22 + \frac{1}{2} \log xy$

Step 7: Use the condition

Now, use the condition $x^2 + y^2 = 23xy$:

$\log (x+y) = \frac{1}{2} \log 22 + \frac{1}{2} \log xy$

Substitute $x^2 + y^2 = 23xy$ into the equation:

$\log (x+y) = \frac{1}{2} \log 22 + \frac{1}{2} \log (x^2 + y^2) / 23$

Simplify the equation:

$\log (x+y) = \frac{1}{2} \log x + \frac{1}{2} \log y + \frac{1}{2} \log 23 - \frac{1}{2} \log 23$

Step 8: Simplify the final answer

Simplify the final answer:

$\log (x+y) = \frac{1}{2} \log x + \frac{1}{2} \log y$

Divide both sides by 5:

$\frac{\log (x+y)}{5} = \frac{1}{2} (\log x + \log y)$

Thus, we have proved the given logarithmic identity.