Problem 20 - Technical Mathematics with Calculus

Question

Solve for \(x\). Give any approximate results to three significant digits. Check your answers. $$2 \log (x+1)=3$$

Step-by-Step Solution

Answer

The solution to the equation is \(x \approx 30.623\).

Step 1: Rewrite the Logarithmic Equation in Exponential Form

Use the fact that if \(a \log(b) = c\), then \(b = 10^{\frac{c}{a}}\). Therefore, \(x+1 = 10^{\frac{3}{2}}\). This step gets rid of the logarithm and turns the equation into one that is more straightforward to solve.

Step 2: Calculate 10 to the Power of 3/2

Calculate \(10^{\frac{3}{2}}\) to get the value that \(x+1\) is equal to. Keep in mind that \(10^{\frac{3}{2}}\) is equivalent to \(\sqrt{10^3}\), or \(10^{1.5}\), which is approximately 31.623.

Step 3: Solve for \(x\)

Subtract 1 from both sides of the equation \(x+1 = 31.623\) to solve for \(x\). Therefore, \(x = 31.623 - 1\).

Step 4: Simplify the Expression to Find the Value of \(x\)

Subtract 1 from 31.623 to find the value of \(x\). Thus, \(x \approx 30.623\).

Step 5: Check the Solution

Substitute \(x \approx 30.623\) back into the original equation and check if the left side equals the right side. Calculate \(2 \log(30.623+1)\) and ensure that this is approximately equal to 3.

Key Concepts

Logarithm to Exponential Form

Understanding how to convert a logarithmic equation to exponential form is a crucial step in solving logarithmic problems. The basic form of a logarithm states that \(\log_b(a) = c\) is equivalent to \(b^c = a\), where \(b\) is the base, \(a\) is the result, and \(c\) is the power to which the base is raised. This conversion is vital because it shifts the equation from a log form, which can be complex to solve, to an exponential form, which is often more straightforward.

Using this conversion, the equation from our exercise can be transformed. Initially, we have \(2 \log (x+1)=3\). By utilizing the conversion process, the equation becomes \(x+1 = 10^{\frac{3}{2}}\), effectively removing the logarithm and setting up the equation for a more direct solution. This step is akin to translating a sentence from a foreign language into your native tongue, making it easier to comprehend and address.

Calculating Powers of 10

Working with powers of 10 is a widespread operation in mathematics, especially in scientific notation. To calculate \(10^\frac{3}{2}\), we understand that any number to the power of \(\frac{1}{2}\) is the square root of that number. Therefore, \(10^\frac{3}{2}\) is the same as the square root of \(10^3\), which is \(10 \times 10 \times 10\).

The computation simplifies to finding the square root of 1000, which is approximately 31.623. To facilitate this, you might rewrite \(10^\frac{3}{2}\) as \(10^{1.5}\), which represents the number 10 to the power of 1 and a half times, confirming that indeed it’s a composite action - multiply 10 by itself once, and then take the square root. By mastering the concept of calculating powers of 10, complex expressions become manageable, and solving equations like those in the exercise becomes almost routine.

Checking Solutions in Logarithms

When solving logarithmic equations, it is essential to check each potential solution to ensure it is valid. This is because logarithms are only defined for positive arguments, and some steps in solving logarithmic equations might introduce extraneous solutions. After finding an approximate value of \(x\), like in our step-by-step solution, where \(x \approx 30.623\), we must verify the solution by plugging it back into the original equation.

To check the solution, take the original equation \(2 \log (x+1)=3\) and substitute \(x\) with 30.623. This gives us \(2 \log(30.623+1)\) or \(2 \log(31.623)\). Upon evaluating this logarithm, we aim to see if it is approximately 3, the right side of the original equation. If the values agree within an accepted margin of error, typically caused by rounding during calculation, then the solution is indeed correct. Through this process, we ensure that our final answer not only solves the equation mathematically but is also meaningful in the context of logarithms.