Question

4^1+log(x+y)=12 log1/3(x+y)+log1/3(x-y)=-1

Answer 1

4^(1+log4(x+y))=12,  

log1/3(x+y)+log1/3(x-y)=-1

4^(1+log4(x+y))=4·4^log4(x+y)=12=4·3

4·4^log4(x+y)=4·3

разделим обе части на 4

4^log4(x+y)=3

(х+у) ^log4(4)=3

после упрощения получили простое уравнение:

х+у=3

log1/3(x+y)+log1/3(x-y)=-1= log 1/3 (3)

получаем:

(х+у) (х-у) =3

подставим сюда (х+у) =3

3(х-у) =3

х-у=3/3=1

получили второе уравнение системы

х-у=1

Получили систему

х+у=3

х-у=1

2х=4; х=2; у=3-х; у=1

Ответ: х=2; у=1

Answer 2

To solve the system of equations, let's start with the first equation:

1. $4^1 + \log(x+y) = 12$

Simplify the left side:

$4 + \log(x+y) = 12$

Now, subtract 4 from both sides:

$\log(x+y) = 8$

Now, rewrite the equation in exponential form:

$x + y = 10^8$

Next, let's move on to the second equation:

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

Combine the logarithms using the product rule:

$\log_{\frac{1}{3}}((x+y)(x-y)) = -1$

Now, rewrite the equation in exponential form:

$(x+y)(x-y) = \left(\frac{1}{3}\right)^{-1}$

$(x+y)(x-y) = 3$

Now, we have a system of equations:

1. $x + y = 10^8$

2. $(x+y)(x-y) = 3$

Let's solve this system. We can use substitution or elimination methods. Let's use substitution:

From equation 1, we can express $x$ as $x = 10^8 - y$.

Now, substitute this value of $x$ into equation 2:

$(10^8 - y + y)(10^8 - y - y) = 3$

$10^8 \cdot (10^8 - 2y) = 3$

$10^{16} - 2 \cdot 10^8 \cdot y = 3$

Now, isolate $y$:

$2 \cdot 10^8 \cdot y = 10^{16} - 3$

$y = \frac{10^{16} - 3}{2 \cdot 10^8}$

$y = \frac{10^{16}}{2 \cdot 10^8} - \frac{3}{2 \cdot 10^8}$

$y = \frac{1}{2} \cdot 10^8 - \frac{3}{2 \cdot 10^8}$

$y = \frac{10^8 - 3}{2 \cdot 10^8}$

$y = \frac{99999997}{2 \cdot 10^8}$

Now, substitute the value of $y$ back into the expression for $x$:

$x = 10^8 - \frac{99999997}{2 \cdot 10^8}$

$x = \frac{2 \cdot 10^8}{2 \cdot 10^8} - \frac{99999997}{2 \cdot 10^8}$

$x = \frac{2 \cdot 10^8 - 99999997}{2 \cdot 10^8}$