Problem 56 - College Algebra Essentials

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. $$\log _{2}(x+50)=5$$

Step-by-Step Solution

Answer

The solution of the equation \(\log _{2}(x+50)=5\) is \(x = -18\), and after the decimal approximation, it's \(x\approx -18.00\).

Step 1: Convert the logarithmic equation into an exponential equation

The logarithmic equation \(\log _{2}(x+50)=5\) is equivalent to the exponential equation \(2^5 = x + 50\). This follows from the definition of a logarithm: if \(\log_b(a) = n\) then \(b^n = a\). In this case, \(b = 2\), \(n = 5\), and \(a = x + 50\). So we can write the equation as \(2^5 = x + 50\).

Step 2: Solve the exponential equation for \(x\)

To solve for \(x\) in the equation \(2^5 = x + 50\), we first calculate two to the fifth power to get 32. So, the equation becomes \(32 = x + 50\). Then, subtract 50 from both sides of the equation to isolate \(x\), giving us \(-18 = x\).

Step 3: Check the solution

The solution \(-18 = x\) needs to be checked to make sure it fits the original logarithmic equation's domain. Because \(x + 50\) is inside a log function, it must always be greater than zero for the logarithm to be defined. Plugging \(-18\) back into the original logarithmic equation gives us \(log_2((-18)+50)\). Since \((-18) + 50\) is positive, the logarithm is defined and our solution for \(x\) is valid.