Question

Xxny satisfy both the equation log x to the base 3 + log y to the base 2 equal to 2 and 3 to the power x - 2 to the power y = 23 then sum of all

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

5

Answer 2

The given system of equations is:

1. $\log_3 x + \log_2 y = 2$
2. $3^x - 2^y = 23$

Step 1: Determine the domain of the variables.

For the logarithms to be defined, $x > 0$ and $y > 0$.

Step 2: Solve by inspection or trial and error for integer solutions.

Let's analyze the second equation: $3^x - 2^y = 23$. We can test integer values for $x$:

• If $x=1$, $3^1 - 2^y = 23 \implies 3 - 2^y = 23 \implies 2^y = -20$. No real solution for $y$.
• If $x=2$, $3^2 - 2^y = 23 \implies 9 - 2^y = 23 \implies 2^y = -14$. No real solution for $y$.
• If $x=3$, $3^3 - 2^y = 23 \implies 27 - 2^y = 23 \implies 2^y = 4$. This implies $y=2$.

So, $(x,y) = (3,2)$ is a potential solution.

Step 3: Verify the potential solution with the first equation.

Substitute $(x,y) = (3,2)$ into the first equation: $\log_3 x + \log_2 y = \log_3 3 + \log_2 2 = 1 + 1 = 2$. The solution $(3,2)$ satisfies both equations.

Step 4: Prove the uniqueness of the solution.

From the first equation, we can express $y$ in terms of $x$: $\log_2 y = 2 - \log_3 x$ $y = 2^{(2 - \log_3 x)}$

Substitute this expression for $y$ into the second equation: $3^x - 2^{(2 - \log_3 x)} = 23$

Let $f(x) = 3^x - 2^{(2 - \log_3 x)}$. We need to determine if $f(x)=23$ has a unique solution. We can do this by examining the derivative of $f(x)$. $f'(x) = \frac{d}{dx}(3^x) - \frac{d}{dx}(2^{(2 - \log_3 x)})$

First term: $\frac{d}{dx}(3^x) = 3^x \ln 3$.

Second term: Let $u = 2 - \log_3 x$. Then $\frac{d}{dx}(2^u) = 2^u \ln 2 \cdot \frac{du}{dx}$. Calculate $\frac{du}{dx}$: $u = 2 - \frac{\ln x}{\ln 3}$ $\frac{du}{dx} = 0 - \frac{1}{\ln 3} \cdot \frac{1}{x} = -\frac{1}{x \ln 3}$.

So, $\frac{d}{dx}(2^{(2 - \log_3 x)}) = 2^{(2 - \log_3 x)} \ln 2 \cdot \left(-\frac{1}{x \ln 3}\right) = -\frac{2^{(2 - \log_3 x)} \ln 2}{x \ln 3}$.

Now, substitute these back into $f'(x)$: $f'(x) = 3^x \ln 3 - \left(-\frac{2^{(2 - \log_3 x)} \ln 2}{x \ln 3}\right)$ $f'(x) = 3^x \ln 3 + \frac{2^{(2 - \log_3 x)} \ln 2}{x \ln 3}$

For $x > 0$:

• $3^x > 0$ and $\ln 3 > 0$, so $3^x \ln 3 > 0$.
• $2^{(2 - \log_3 x)} > 0$, $\ln 2 > 0$, $x > 0$, and $\ln 3 > 0$, so $\frac{2^{(2 - \log_3 x)} \ln 2}{x \ln 3} > 0$.

Since both terms are positive, $f'(x) > 0$ for all $x > 0$. This means $f(x)$ is a strictly increasing function. A strictly increasing function can intersect a horizontal line (like $f(x)=23$) at most once. Since we found one solution $(x,y) = (3,2)$, this must be the unique solution to the system.

Step 5: Calculate the requested sum.

The question asks for "sum of all". Given that there is a unique solution $(x,y)=(3,2)$, it is highly probable that it refers to the sum of $x$ and $y$. Sum $= x+y = 3+2 = 5$.