Problem 58 - Algebra and Trigonometry

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{7}(x+2)=-2 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -97/49\) or \(x = -1.98\) (approx), and it is a valid solution.

Step 1: Convert logarithm to exponential form

The logarithm equation \( \log_{7} (x + 2) = -2 \) can be converted into an exponential form using the formula \(b^{y}=x\), where \(b\) is the base, \(y\) is the exponent and \(x\) is the result. Thus, the equation becomes \(7^{-2} = x + 2\).

Step 2: Simplify the equation

Solving the equation \(7^{-2} = x + 2\), we obtain the value of \(x\). To do this, firstly calculate \(7^{-2}\) to get \(1/49\). Then subtract 2 from both sides of the equation leading to \(x = 1/49 - 2\). To ease further calculation, 2 can be converted to \(98/49\). Thus, the equation becomes \(x = 1/49 - 98/49\).

Step 3: Solve for \(x\)

Now subtract the fractions on the right side of the equation to get the value for \(x\), i.e., \(x = -97/49\). Configuring this to a decimal value gives \(x = -1.98\) (rounded to two decimal places).

Step 4: Verify the solution

After the calculation step, it is necessary to verify the solution against the original logarithmic equation. In this case, \(x = -1.98\) cannot serve as a solution as it does not fall within the domain of the original logarithmic expression. This is because the expression inside the logarithm (i.e., \(x + 2\)) should be greater than zero or the logarithm is undefined. In this case, with \(x = -1.98\), the expression \(x + 2 = -1.98 + 2 = 0.02\) which is greater than zero, so it is valid.

Key Concepts

Logarithm to Exponential Form

Understanding how to convert logarithmic equations to exponential form is a fundamental skill in algebra. The general logarithmic equation \(\log_b(x) = y\) can be rewritten in exponential form as \(b^y = x\). This transformation allows you to approach the equation from a different perspective. For instance, if you have the logarithmic equation \(\log_7(x+2) = -2\), converting it to exponential form changes it to \(7^{-2} = x + 2\). This step is critical and provides a clear pathway toward finding the value of \(x\).

When you're faced with a logarithmic equation, remember to first rewrite it in the exponential form to make the equation more workable. This strategy simplifies the problem and prepares you for further calculations to solve for the unknown variable.

Domain of Logarithmic Functions

The domain of a function is the set of all possible input values (typically \(x\) values) that the function can accept without leading to any undefined or non-real results. For logarithmic functions, this is crucial as the argument of the logarithm— the value inside the logarithm — must be greater than zero. Why is this important? Because you cannot take a logarithm of a non-positive number in real numbers.

In our example, the domain of \(\log_7(x+2)\) demands that \(x + 2 > 0\), meaning \(x > -2\). Any solution outside this interval is invalid, as it would not satisfy the initial conditions of the logarithmic function. This is a step you must not overlook after solving the equation, to ensure the validity of your solution.

Exponential and Logarithmic Properties

Exponential and logarithmic properties are essential tools that simplify the process of solving equations involving exponents and logarithms. Some key properties to remember include:

• The product property: \(\log_b(mn) = \log_b(m) + \log_b(n)\)
• The quotient property: \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\)
• The power property: \(\log_b(m^n) = n\cdot \log_b(m)\)
• Change of base formula: \(\log_b(m) = \frac{\log_k(m)}{\log_k(b)}\) where \(k\) is a positive real number.

These properties enable you to manipulate logarithmic expressions and facilitate the transition between logarithmic and exponential forms. They are vital when simplifying logarithmic expressions before solving the equations and can also assist in verifying the correctness of your solutions.

Verifying Solutions of Logarithmic Equations

Finding a solution to a logarithmic equation is part of the process, but verification is just as important to ensure accuracy. To verify a solution, you need to plug it back into the original logarithmic equation and see if it results in a true statement. Also, check if the solution falls within the domain of the logarithmic function.

Take our example, after solving \(\log_7(x+2) = -2\) and finding \(x = -1.98\). To verify, when \(x + 2\) is plugged back into the logarithm, it should result in \(\log_7(0.02)\), which must equal \(\log_7(7^{-2})\) or \(\log_7(\frac{1}{49})\), thereby validating our solution — but only if \(x + 2 > 0\). In this case, since \(0.02 > 0\), the solution is indeed valid. Always perform this verification step to confirm the solutions found are true and within the allowable domain.