If logx 27+ log y 4 = 5 and Logx 27-log y 4=1, what is X and Y?

To solve the equations logx 27 + logy 4 = 5 and logx 27 – logy 4 = 1 for the values of X and Y, we’ll use logarithmic properties and algebraic techniques.

Let’s start with the first equation: logx 27 + logy 4 = 5

We know that the sum of logarithms with the same base is equal to the logarithm of their product: logx (27 * 4) = 5

Now, simplify the equation further: logx 108 = 5

To get rid of the logarithm, we can rewrite it in exponential form: x^5 = 108

Now, let’s move on to the second equation: logx 27 – logy 4 = 1

Using the properties of logarithms again, we can subtract the second logarithm from the first one: logx (27 / 4) = 1

Simplify: logx 6.75 = 1

Now, rewrite it in exponential form: x^1 = 6.75

Now, we have two equations:

  1. x^5 = 108
  2. x = 6.75

Let’s solve the second equation for X: x = 6.75

Now, we can substitute this value of X into the first equation: (6.75)^5 = 108

Using a calculator: 6.75 raised to the power of 5 is approximately equal to 2438.83.

So, X is approximately equal to 2438.83.

Now, let’s find Y. We can use the second equation we derived earlier: x = 6.75

Substitute this value into the second equation: 6.75 = 6.75

This equation tells us that Y can be any real number because it doesn’t depend on Y. Therefore, there are infinite solutions for Y.

In summary: X is approximately equal to 2438.83, and Y can be any real number.

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