Problem 42 - Precalculus : Building Concepts and Connections

Question

Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log _{5}(x+3)=1$$

Step-by-Step Solution

Answer

Hence, the solution to the equation \(\log _{5}(x+3)=1\) is \(x = 2\).

Step 1: Write the equation in exponential form

The given equation is \(\log _{5}(x+3)=1\). We can rewrite this as an exponential equation. Using the definition of logarithm, we can write this into an equivalent exponential form as \(5^1 = x+3\).

Step 2: Solve for x

Solving the equation \(5 = x+3\) for x, we subtract 3 from both sides of the equation to isolate x, therefore the solution is \(x = 5-3 = 2\).

Step 3: Check for extraneous solutions

Substitute \(x = 2\) back into the original equation to confirm it doesn't result in an argument of the logarithm that is less than zero, as logarithms are undefined for values less than or equal to zero. The resulting equation \(\log _{5}(2+3)\) simplifies to \(\log _{5}(5)\) which is indeed 1, thus confirming that \(x=2\) is a valid solution rather than an extraneous one.

Key Concepts

Exponential Form

The transformation of a logarithmic equation into exponential form is a crucial step in solving logarithmic problems. This conversion is based on the fundamental relationship between logarithms and exponents. For the given exercise, \(\log _{5}(x+3)=1\), the exponential form is obtained by recognizing that the logarithm equation is essentially asking, 'To what power must the base (5) be raised to produce the number (x+3)?'. The answer to that question is the number on the other side of the equation, which is 1. Hence, we express this relationship exponentially as \(5^1 = x+3\).

Understanding this conversion process is essential, as it is the first step towards isolating the variable and finding the solution to the logarithmic equation. A logarithm is essentially an exponent in another form, and this connection allows us to switch between logarithmic and exponential forms to simplify the problem solving process.

Logarithm Definition

The concept of a logarithm is fundamental to understanding how to manipulate and solve logarithmic equations. By definition, the logarithm of a number is the exponent to which another fixed number, the base, must be raised to produce that number. It is the inverse operation to exponentiation. In mathematical terms, if \(b^y = x\), then \(\log_b(x) = y\), where \(b\) is the base, \(x\) is the number, and \(y\) is the logarithm or exponent.

For example, in our exercise, the logarithm \(\log _{5}(x+3)\) asks for the power to which the base 5 must be raised to equal \(x+3\). When we say \(\log _{5}(x+3)=1\), we mean that 5 raised to the power of 1 equals \(x+3\). This relationship is essential in transferring a logarithmic statement into one involving exponents, thereby simplifying the solution process.

Extraneous Solutions

In solving equations, particularly those involving radical or logarithmic expressions, we sometimes come across solutions that do not actually satisfy the original equation; these are known as extraneous solutions. They frequently arise from the steps taken during the algebraic manipulation of the equation, such as squaring both sides of an equation or applying logarithmic principles.

Therefore, it is always necessary to check potential solutions back in the original equation, as was done in step 3 of our solution process. In the exercise \(\log _{5}(x+3)=1\), after solving for \(x\) we find that \(x=2\). To ensure this is not an extraneous solution, we substitute \(x=2\) back into the original equation, resulting in \(\log _{5}(5)\), which simplifies correctly to 1, the same value as the original right-hand side of the equation, confirming that \(x=2\) is indeed a valid solution, not an extraneous one. It is crucial to make this verification step a standard practice to ensure the accuracy of the solutions derived from equations, especially logarithmic ones.