Question

solve the following logarithmic equation. be sure to reject any value of x that is not in the domain of the original logarithmic expression. give the exact answer. \\(\log_{2}(\sqrt{x} + \sqrt{3}) = 6\\) solve the equation. select the correct choice below and, if necessary, fill in the answer box to complete your choice. \\(\bigcirc\\) a. the solution set is \\(\\{\\}\\). (type an integer or a simplified fraction.) \\(\bigcirc\\) b. there are infinitely many solutions. \\(\bigcirc\\) c. there is no solution.

Explanation:

Step1: Convert log to exponential form

Given \(\log_{2}(x + \sqrt{3}) = 6\), by the definition of logarithms, if \(\log_{b}a = c\), then \(a = b^{c}\). So we have \(x+\sqrt{3}=2^{6}\).

Step2: Calculate \(2^{6}\) and solve for \(x\)

Calculate \(2^{6}=64\). Then the equation becomes \(x+\sqrt{3}=64\). Subtract \(\sqrt{3}\) from both sides: \(x = 64-\sqrt{3}\). Now check the domain: the argument of the logarithm \(x+\sqrt{3}\) must be positive. For \(x = 64-\sqrt{3}\), \(x+\sqrt{3}=64>0\), so it is in the domain.

Answer:

The solution set is \(\{64 - \sqrt{3}\}\) (corresponding to option A with the value \(64 - \sqrt{3}\))