Question3. If $\left(\frac{1}{3}\right)^{x+y}=243$ and $\log _{2}(2 y-x)=2$ , calculate the values of $x$ and $y$ algebraically.

Studdy Solution

STEP 1

1. The equation $\left(\frac{1}{3}\right)^{x+y} = 243$ is an exponential equation.
2. The equation $\log_{2}(2y-x) = 2$ is a logarithmic equation.
3. We will need to solve these equations simultaneously to find the values of $x$ and $y$ .

STEP 2

1. Solve the exponential equation for $x+y$ .
2. Solve the logarithmic equation for $2y-x$ .
3. Use the results from the first two steps to solve for $x$ and $y$ .

STEP 3

First, rewrite the exponential equation $\left(\frac{1}{3}\right)^{x+y} = 243$ using a common base. Note that $243 = 3^5$ , so we can rewrite the equation as:
$\left(\frac{1}{3}\right)^{x+y} = 3^{-1(x+y)} = 3^5$
Equating the exponents, we have:
$-1(x+y) = 5$
Solving for $x+y$ , we get:
$x+y = -5$

STEP 4

Next, solve the logarithmic equation $\log_{2}(2y-x) = 2$ . By the definition of a logarithm, this equation implies:
$2y-x = 2^2$
Simplifying, we find:
$2y-x = 4$

STEP 5

Now, we have the system of equations:
1. $x+y = -5$
2. $2y-x = 4$
Add the two equations to eliminate $x$ :
$(x+y) + (2y-x) = -5 + 4$
Simplifying, we get:
$3y = -1$
Solve for $y$ :
$y = -\frac{1}{3}$
Substitute $y = -\frac{1}{3}$ back into the first equation $x+y = -5$ :
$x - \frac{1}{3} = -5$
Solving for $x$ , we find:
$x = -5 + \frac{1}{3}$ $x = -\frac{15}{3} + \frac{1}{3}$ $x = -\frac{14}{3}$
The values of $x$ and $y$ are:
$x = -\frac{14}{3}, \quad y = -\frac{1}{3}$

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