Let (x₀, y₀) be the solution of the following equations (2x)^(ln2) = (3y)^(ln3), 3^(lnx) = 2^(lny).

Q: Wait, why can't we just take the log of the whole equation directly? Like $\ln((2x)^{\ln 2}) = \ln((3y)^{\ln 3})$?

You absolutely can! In fact, that is exactly what we did. The trap students often fall into is trying to distribute the log across addition or subtraction. Remember, $\ln(A+B)$ is not $\ln A + \ln B$. Always keep the arguments of your logarithms intact until you are ready to use the product rule $\ln(ab) = \ln a + \ln b$.

Q: Why did we substitute $\ln y$ instead of solving for $y$ directly?

Solving for $y$ directly would involve exponentiating, which leads to messy expressions like $y = x^{\frac{\ln 3}{\ln 2}}$. By working in 'log-space' (treating $\ln x$ and $\ln y$ as variables), we keep the algebra linear and clean. It is the difference between walking on a flat road versus climbing a mountain.

Q: Is it possible to have a negative solution for $x$?

Great question! Look at the original equation: $(2x)^{\ln 2}$. The base of an exponential function must be positive for the function to be well-defined in the real number system. Therefore, $2x > 0$, which implies $x > 0$. Always check the domain of your logarithmic and exponential functions before accepting your final answer.

Question

Let $(x_0, y_0)$ be the solution of the following equations $(2x)^{\ln 2} = (3y)^{\ln 3}, 3^{\ln x} = 2^{\ln y}$. Then $x_0$ is

Options:

(A) 1/6

(B) 1/3

(C) 1/2

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given system of equations is:
$$(2x)^{\ln 2} = (3y)^{\ln 3}$$
$$3^{\ln x} = 2^{\ln y}$$

Our goal is to liberate the variables $x$ and $y$ from their exponential positions by utilizing the properties of logarithms.

### The Power of the Logarithm

We begin by applying the natural logarithm, $\ln$, to both sides of the second equation:
$$3^{\ln x} = 2^{\ln y}$$

Using the power property $\ln(a^b) = b \ln a$, we bring the exponents down:
$$(\ln x)(\ln 3) = (\ln y)(\ln 2)$$

We can now define the relationship between $\ln y$ and $\ln x$ as:
$$\ln y = \frac{(\ln x)(\ln 3)}{\ln 2}$$

### The Trap of the Product

Next, we apply the logarithm to the first equation:
$$(2x)^{\ln 2} = (3y)^{\ln 3}$$
$$(\ln 2) \ln(2x) = (\ln 3) \ln(3y)$$

Applying the product property $\ln(ab) = \ln a + \ln b$, we expand the expression:
$$(\ln 2)(\ln 2 + \ln x) = (\ln 3)(\ln 3 + \ln y)$$

### The Grand Convergence

We substitute our expression for $\ln y$ from the previous phase into this equation:
$$(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3) \left( \frac{(\ln x)(\ln 3)}{\ln 2} \right)$$

Expanding and grouping the terms involving $\ln x$ on one side, we obtain:
$$(\ln x) \left( \ln 2 - \frac{(\ln 3)^2}{\ln 2} \right) = (\ln 3)^2 - (\ln 2)^2$$

To simplify the left side, we find a common denominator:
$$(\ln x) \left( \frac{(\ln 2)^2 - (\ln 3)^2}{\ln 2} \right) = (\ln 3)^2 - (\ln 2)^2$$

### Final Calculation

Notice that $(\ln 3)^2 - (\ln 2)^2 = -((\ln 2)^2 - (\ln 3)^2)$. Dividing both sides by the common term $((\ln 2)^2 - (\ln 3)^2)$, we get:
$$\frac{\ln x}{\ln 2} = -1$$

This simplifies to $\ln x = -\ln 2$, which is equivalent to $\ln x = \ln(2^{-1})$. Therefore, the final value is:
**$x = \frac{1}{2}$**

Animated Solution for Mathematics - Basic Mathematics: Let $(x_{0}, y_{0})$ be the solution of the following equations $(2 x)^{l n 2} = (3 y)^{l n 3}, 3^{l n x} = 2^{l n y}$ . Then $x_{0}$ is

Select Answer:

Visualized Solution

Problem Analysis

Transforming Equation 2

Isolating $ln y$

Transforming Equation 1

Expanding Logarithms

Substitution Step

Expanding and Grouping

Factoring $ln x$

Solving for $ln x$

Final Answer

The Sigma Insight: Logarithmic Equations and Inequalities

Analyzing the Setup

The Power of the Logarithm

The Trap of the Product

The Grand Convergence

Final Calculation

Similar Questions

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If $3^{x} = 4^{x - 1}$ , then $x =$

The number of integral solution $x$ of $lo g_{(x + \frac{7}{2})} (\frac{x - 7}{2 x - 3})^{2} \geq 0$ is

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The solution of the equation $lo g_{7} lo g_{5} (x + 5 + x) = 0$ is \dots.

If $lo g_{0.3} (x - 1) < lo g_{0.09} (x - 1)$ , then $x$ lies in the interval

Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (x0​,y0​) be the solution of the following equations (2x)ln2=(3y)ln3,3lnx=2lny. Then x0​ is

Select Answer:

Visualized Solution

Problem Analysis

Transforming Equation 2

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } lny

Transforming Equation 1

Expanding Logarithms

Substitution Step

Expanding and Grouping

Factoring .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } lnx

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } lnx

Final Answer

The Sigma Insight: Logarithmic Equations and Inequalities

Analyzing the Setup

The Power of the Logarithm

The Trap of the Product

The Grand Convergence

Final Calculation

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