Question

4^1+ log4(x+y)=12 log1/3(x+y)+ log1/3 (x-y)=-1Это система уравнений

Answer 1

В приложении .............

Answer 2

Yes, the given equations form a system of equations. Let's solve the system step-by-step.

The system of equations is:

1. $4^1 + \log_4(x+y) = 12$
2. $\log_{\frac{1}{3}}(x+y) + \log_{\frac{1}{3}}(x-y) = -1$

Step 1: Simplify the equations

1. $4^1 = 4$
2. $\log_{\frac{1}{3}}(x+y) + \log_{\frac{1}{3}}(x-y) = -1$

Step 2: Apply logarithm properties Recall that $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$.

Using this property in equation 2: $\log_{\frac{1}{3}}((x+y) \cdot (x-y)) = -1$

Step 3: Convert the logarithmic equation to exponential form Recall that $\log_a(b) = c$ is equivalent to $a^c = b$.

Using this in equation 2: $\frac{1}{3}^{-1} = (x+y) \cdot (x-y)$

Step 4: Simplify $\frac{1}{3}^{-1} = \frac{1}{\frac{1}{3}} = 3$

So, we have: $3 = (x+y) \cdot (x-y)$

Step 5: Solve for x and y Now, we have a system of equations:

1. $4 + \log_4(x+y) = 12$
2. $3 = (x+y) \cdot (x-y)$

Let's solve equation 1 first: $4 + \log_4(x+y) = 12$

Subtract 4 from both sides: $\log_4(x+y) = 8$

Now, rewrite this in exponential form: $4^8 = x+y$

$x+y = 65536$

Next, let's solve equation 2: $3 = (x+y) \cdot (x-y)$

Substitute the value we found for $x+y$: $3 = 65536 \cdot (x-y)$

Now, solve for $x-y$: $x-y = \frac{3}{65536}$

Now, we have a system of equations:

1. $x+y = 65536$
2. $x-y = \frac{3}{65536}$

Add the two equations to eliminate y: $2x = 65536 + \frac{3}{65536}$

Now, solve for x: $x = \frac{65536 + \frac{3}{65536}}{2}$

Calculate the value of x, then substitute it back into one of the equations to solve for y. The solution will be the values of x and y that satisfy both equations in the system.