Problem 54 - Introductory and Intermediate Algebra for College Students 4th

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. $$\log _{2}(4 x+1)=5$$

Step-by-Step Solution

Answer

The solution to the equation \(\log _{2}(4 x+1)=5\) is \(x=7.75\).

Step 1: Convert logarithm into exponential form

The conversion takes the form \(b^y=x\), where \(b\) is the base, \(y\) is the exponent, and \(x\) is the log value. In the equation \(\log _{2}(4 x+1)=5\), \(2\) is the base, \(5\) is the exponent, and \(4x+1\) is the log value. Therefore, the logarithmic equation \(\log _{2}(4 x+1)=5\) becomes \(2^5 = 4x+1\). So, \(32= 4x+1\).

Step 2: Solve for \(x\)

Subtract 1 from both sides of the equation to isolate \(4x\). So, \(32-1=4x\), which simplifies to \(31=4x\). Then divide both sides by 4, so \(x = 31/4\).

Step 3: Evaluate Decimal Approximation

If required, a decimal approximation could be done by dividing 31 with 4 using a calculator. It gives \(x\) approximately equal to 7.75 to two decimal places.

Step 4: Checking Solution for Validity

Substitute the value of \(x\) back into the original equation to see if it holds true. After substituting \(7.75\), we get \(\log _{2}(4(7.75)+1)=5\), which simplifies to \(\log _{2}(32)=5\), and this is true. Therefore, \(x=7.75\) is the solution.