Logarithm Equation: Solve For Y

When faced with a logarithmic equation like log 5 + log y = log 40, the immediate goal is to isolate the variable, in this case, $y$. This type of problem often appears in mathematics, particularly in algebra and pre-calculus, and understanding how to manipulate logarithmic properties is key to finding the solution. The core principle we'll employ here relies on the fundamental properties of logarithms, specifically the product rule. This rule states that the sum of logarithms with the same base is equal to the logarithm of the product of their arguments. In mathematical terms, this is expressed as $\log_b(M) + \log_b(N) = \log_b(MN)$. Applying this to our equation, we can combine the terms on the left side. The equation log 5 + log y = log 40 becomes log (5 * y) = log 40. Notice that the 'log' on both sides implies a common base, usually base 10 if not specified, or it could be the natural logarithm (ln) with base $e$. Regardless of the base, as long as it's consistent, the property holds true. Once we have a single logarithm on each side of the equation, we can proceed to eliminate the logarithms. If $\log_b(A) = \log_b(B)$, then it logically follows that $A = B$. This is because the logarithm function is a one-to-one function. Therefore, from log (5y) = log 40, we can directly equate the arguments: 5y = 40. This transforms our logarithmic equation into a simple linear equation, which is much easier to solve for $y$. To isolate $y$, we simply divide both sides of the equation by 5. So, $y = 40 / 5$, which gives us $y = 8$. It's always a good practice to check our solution by substituting it back into the original equation to ensure it holds true and that the arguments of the logarithms remain positive (as the logarithm of a non-positive number is undefined in the real number system). Substituting $y=8$ into log 5 + log y = log 40, we get log 5 + log 8 = log 40. Using the product rule again, log (5 * 8) = log 40, which simplifies to log 40 = log 40. This confirms that our solution $y=8$ is correct. Mastering these basic logarithmic properties is fundamental for tackling more complex equations and understanding exponential relationships in various mathematical and scientific contexts.

Understanding the Properties of Logarithms

To effectively solve the equation log 5 + log y = log 40, a solid grasp of the fundamental properties of logarithms is essential. These properties allow us to simplify and manipulate logarithmic expressions, making complex equations more manageable. The property most relevant here is the product rule, which states that for any positive numbers $M$ and $N$, and any base $b$ (where $b > 0$ and $b \neq 1$), the sum of their logarithms is equal to the logarithm of their product: $\log_b(M) + \log_b(N) = \log_b(MN)$. In our equation, we have log 5 and log y. Assuming these are common logarithms (base 10), we can apply the product rule to the left side of the equation: $\log(5) + \log(y) = \log(5 imes y) = \log(5y)$. So, the original equation log 5 + log y = log 40 transforms into log(5y) = log 40. Another crucial property is the quotient rule, which states $\log_b(M) - \log_b(N) = \log_b(M/N)$. While not directly used in this specific problem's simplification, it's part of the logarithmic toolkit. Similarly, the power rule, $\log_b(M^p) = p \log_b(M)$, is invaluable for dealing with exponents within logarithms. Furthermore, understanding the one-to-one property of logarithms is critical for solving equations. This property asserts that if $\log_b(A) = \log_b(B)$, then it must be true that $A = B$. This is because the logarithmic function is strictly increasing (or decreasing) and therefore can only take on a specific value once. In our simplified equation, log(5y) = log 40, we have the same logarithmic function (log) on both sides. Applying the one-to-one property, we can equate the arguments: $5y = 40$. This step is what bridges the gap from a logarithmic equation to a simple algebraic one. Before we even start solving, it's vital to consider the domain of logarithmic functions. The argument of a logarithm must always be positive. In our original equation, this means $y$ must be greater than 0 ($y > 0$) for log y to be defined. If our solution yields a non-positive value for $y$, it would be an extraneous solution and must be discarded. In this case, our solution $y=8$ is indeed positive, so it's a valid solution. Familiarity with these properties allows us to not only solve equations like this efficiently but also to simplify complex expressions and prove other mathematical identities related to logarithms.

Step-by-Step Solution Process

Let's break down the process of solving the logarithmic equation log 5 + log y = log 40 into clear, actionable steps. Our primary objective is to isolate the variable $y$. The first step involves utilizing the properties of logarithms to simplify the equation. Specifically, we apply the product rule of logarithms, which states that $\log_b(M) + \log_b(N) = \log_b(MN)$. In our equation, the base of the logarithm is not explicitly stated, so we assume it's the common logarithm (base 10). Applying the product rule to the left side of the equation, log 5 + log y, we combine these two terms into a single logarithm: $\log(5 \times y)$, which simplifies to $\log(5y)$. Now, our equation becomes log(5y) = log 40. The next crucial step leverages the one-to-one property of logarithms. This property states that if $\log_b(A) = \log_b(B)$, then $A = B$. Since we have the same logarithmic function on both sides of our equation, we can equate their arguments. Therefore, from log(5y) = log 40, we deduce that 5y = 40. At this point, the equation has been transformed from a logarithmic equation into a simple linear equation. The final step is to solve this linear equation for $y$. To isolate $y$, we perform the inverse operation of multiplication, which is division. We divide both sides of the equation 5y = 40 by 5: $y = 40 / 5$. Calculating the result, we find that $y = 8$. It is always a good practice to verify the solution by substituting it back into the original equation. This ensures that the solution is valid and that no extraneous solutions were introduced. Substituting $y = 8$ into log 5 + log y = log 40, we get log 5 + log 8 = log 40. Using the product rule again on the left side, we have $\log(5 \times 8) = \log 40$. This simplifies to log 40 = log 40, which is a true statement. Additionally, we must ensure that the arguments of the logarithms in the original equation are positive. For log y, the argument is $y$. Our solution $y=8$ is positive, satisfying this condition. Therefore, the solution $y=8$ is correct and valid. This systematic approach, combining logarithmic properties with algebraic manipulation, is fundamental for solving a wide range of logarithmic equations.

Verifying the Solution and Domain Considerations

After solving the logarithmic equation log 5 + log y = log 40 and arriving at a potential solution, $y=8$, the critical next step is to verify the solution. This verification process serves two main purposes: confirming the accuracy of our algebraic manipulation and ensuring that the solution adheres to the domain restrictions inherent in logarithmic functions. To verify the accuracy, we substitute $y=8$ back into the original equation: log 5 + log 8 = log 40. We then apply the properties of logarithms to the left side. Using the product rule ($\log M + \log N = \log(MN)$), we combine log 5 and log 8 into a single logarithm: $\log(5 \times 8)$, which equals log 40. So, the equation becomes log 40 = log 40. Since this statement is true, our calculated value of $y=8$ is mathematically correct within the structure of the equation. Beyond simple algebraic correctness, we must also consider the domain of logarithmic functions. The logarithm of a number is only defined for positive arguments. In the equation log 5 + log y = log 40, we have log y. For log y to be defined in the real number system, the argument, $y$, must be strictly greater than zero ($y > 0$). If our solution had resulted in a value of $y$ that was zero or negative, it would be considered an extraneous solution and would need to be discarded. In this specific problem, our solution $y=8$ is positive. Therefore, it is a valid solution that satisfies the domain requirement for log y. This validation step is crucial, especially in more complex logarithmic equations where multiple solutions might arise from the algebraic process, but only some are valid within the defined domain. For instance, if we were solving an equation like $\log(y^2) = \log(16)$, algebraically we might get $y^2 = 16$, leading to $y=4$ and $y=-4$. However, the original equation has log(y^2), meaning $y^2$ must be positive, which is true for both $y=4$ and $y=-4$. But if the equation were $\log(y) + \log(y) = \log(16)$, then $y$ must be positive, making $y=-4$ an extraneous solution. Always remembering to check the domain of the original logarithmic expressions is key to finding the correct and complete set of solutions. For a deeper dive into logarithmic properties and equations, you can explore resources like Paul's Online Math Notes.