Solve $2^{3x-2}=5^x$: The Definitive Guide

When faced with exponential equations where the bases are different, like the intriguing problem of solving $2^{3x-2}=5^x$, it might initially seem like a puzzle with no easy answer. However, mathematics offers elegant tools to tackle such challenges. The key lies in manipulating the equation using logarithms. Logarithms are the inverse operations of exponentiation, meaning if $b^y = x$, then $\log_b x = y$. This fundamental property allows us to bring down exponents, transforming a complex exponential equation into a more manageable algebraic one. For our specific problem, $2^{3x-2}=5^x$, we have different bases (2 and 5) and exponents that contain the variable 'x'. Our goal is to isolate 'x'. The most effective strategy here is to take the logarithm of both sides of the equation. You can use any base for the logarithm, but the natural logarithm (ln, base e) or the common logarithm (log, base 10) are typically the most convenient choices due to their widespread use and calculator accessibility. By applying the same logarithm to both sides, we maintain the equality of the equation. This step is crucial and forms the foundation for solving this type of problem. Remember, the property of logarithms we'll be using extensively is $\log(a^b) = b \log a$. This rule allows us to convert powers into multipliers, which is exactly what we need to do to get 'x' out of the exponents.

Let's dive into the process of solving the equation $2^{3x-2}=5^x$ using logarithms. First, we apply the natural logarithm (ln) to both sides of the equation. This gives us: $\ln(2^{3x-2}) = \ln(5^x)$. Now, using the logarithm property $\log(a^b) = b \log a$, we can bring the exponents down as multipliers. Applying this to our equation, we get: $(3x-2)\ln 2 = x \ln 5$. At this stage, the equation no longer has 'x' in the exponents, which is a significant simplification. The next step involves algebraic manipulation to isolate 'x'. We need to expand the left side of the equation by distributing $\ln 2$: $3x \ln 2 - 2 \ln 2 = x \ln 5$. Now, we want to gather all terms containing 'x' on one side of the equation and all constant terms on the other. Let's move the $x \ln 5$ term to the left side and the $-2 \ln 2$ term to the right side. This results in: $3x \ln 2 - x \ln 5 = 2 \ln 2$. To further isolate 'x', we can factor it out from the terms on the left side: $x(3 \ln 2 - \ln 5) = 2 \ln 2$. The final step to solve for 'x' is to divide both sides by the coefficient of 'x', which is $(3 \ln 2 - \ln 5)$. This yields the solution: $x = \frac{2 \ln 2}{3 \ln 2 - \ln 5}$. This methodical approach, leveraging the power of logarithms and basic algebraic skills, allows us to find the exact solution to what initially appears to be a complex exponential equation. It's a testament to the systematic nature of mathematics in providing clear paths to solutions.

Let's re-examine the solution we derived for the equation $2^{3x-2}=5^x$, which is $x = \frac{2 \ln 2}{3 \ln 2 - \ln 5}$. This expression represents the exact value of 'x' that satisfies the original equation. Often in mathematics, especially in standardized tests or when precision is paramount, you'll be asked for the exact solution rather than a decimal approximation. This exact form, using logarithms, is the most accurate representation. Now, let's consider the provided options to see which one matches our result. The options are:

A. $\frac{3 \ln 2+\ln 5}{2 \ln 2}$ B. $\frac{2 \ln 2}{3 \ln 2+\ln 5}$ C. $\frac{3 \ln 2-\ln 5}{2 \ln 2}$ D. $\frac{2 \ln 2}{3 \ln 2-\ln 5}$

Comparing our derived solution, $x = \frac{2 \ln 2}{3 \ln 2 - \ln 5}$, directly with the given options, we can see that option D is an exact match. This confirms that our step-by-step application of logarithmic properties and algebraic manipulation was correct. It's important to be comfortable with these logarithmic rules, particularly the power rule $\log(a^b) = b \log a$, and the change of base formula if needed, although in this case, the natural logarithm sufficed. Understanding these properties is fundamental for anyone looking to master exponential and logarithmic equations. They allow us to transform problems into forms that are much easier to solve. For instance, the difference of logarithms, $\ln a - \ln b$, can also be written as $\ln(\frac{a}{b})$. So, the denominator $3 \ln 2 - \ln 5$ could be rewritten as $\ln(2^3) - \ln 5 = \ln 8 - \ln 5 = \ln(\frac{8}{5})$. Thus, an alternative exact form of the solution is $x = \frac{\ln 4}{\ln(8/5)}$. While this is also correct, the form provided in option D is the one typically obtained directly from the method described, and it's the one we should look for when matching against multiple-choice answers. Always double-check your algebraic steps to avoid errors, especially when dealing with signs and rearranging terms. The confidence in your answer grows with each correctly applied rule and each verified step.

To further solidify our understanding of solving exponential equations like $2^{3x-2}=5^x$, it's beneficial to review the core principles and potential pitfalls. The fundamental step involves using logarithms to bring the exponents down. This is only possible when you apply the logarithm to both sides of the equation, ensuring the equality remains valid. The property $\log(a^b) = b \log a$ is your most powerful tool here. Once the exponents are linearized, the problem transforms into a standard algebraic equation. This means you'll be dealing with terms involving 'x' and constant terms. The objective is to group all 'x' terms on one side and all constants on the other. Factoring 'x' out is a common technique to isolate it. For example, if you reach a stage like $Ax + Bx = C$, you factor it to $x(A+B) = C$, and then solve for $x = \frac{C}{A+B}$. Similarly, if you have terms like $Ax - Bx = C$, it becomes $x(A-B) = C$, leading to $x = \frac{C}{A-B}$. In our specific case, after distributing, we had $3x \ln 2 - 2 \ln 2 = x \ln 5$. Rearranging gave us $3x \ln 2 - x \ln 5 = 2 \ln 2$. Factoring out 'x' yielded $x(3 \ln 2 - \ln 5) = 2 \ln 2$. The final step, dividing by the coefficient of 'x', resulted in $x = \frac{2 \ln 2}{3 \ln 2 - \ln 5}$. A potential area for error is in the rearrangement of terms. Be meticulous with signs. Moving a term from one side to another changes its sign. Another common mistake might be misapplying logarithm properties, for instance, confusing $\ln(a-b)$ with $\ln a - \ln b$. Remember that $\ln a - \ln b = \ln(\frac{a}{b})$, but there is no simple rule for $\ln(a-b)$. Always write down the properties clearly and apply them carefully. The choice of logarithm base (natural log or common log) usually doesn't affect the final exact answer, as the base can be converted using the change of base formula: $\log_b a = \frac{\log_c a}{\log_c b}$. However, using 'ln' or 'log' often simplifies notation. When comparing your result to multiple-choice options, ensure that the forms are equivalent. Sometimes, logarithmic expressions can be simplified or rewritten using different properties, so you might need to manipulate your answer or the options to find a match. For instance, $3 \ln 2$ can be written as $\ln(2^3)$ or $\ln 8$. This can help in simplifying the expression or matching it with an option that uses this form. Practicing a variety of these problems will build your confidence and speed in solving them accurately. Remember, the goal is to systematically break down the complex exponential structure into a solvable algebraic form.

In conclusion, solving exponential equations where the bases are different, such as $2^{3x-2}=5^x$, requires a strategic application of logarithmic properties. The process involves taking the logarithm of both sides, using the power rule to bring down exponents, and then employing algebraic manipulation to isolate the variable 'x'. We found the exact solution to be $x = \frac{2 \ln 2}{3 \ln 2 - \ln 5}$, which corresponds to Option D. Mastering these techniques is essential for advanced algebra and calculus. For further exploration into the properties of logarithms and exponential functions, you can refer to resources like Khan Academy's logarithm section or Paul's Online Math Notes on Exponential and Logarithmic Equations.