Problem 99 - Algebra and Trigonometry

Question

Solve each equation. $$ \ln (2 x+1)+\ln (x-3)-2 \ln x=0 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x=1\).

Step 1: Use properties of logarithms

Using the property \(\ln(a)+\ln(b)=\ln(ab)\) and \(\ln(c)-\ln(d)=\ln(\frac{c}{d})\), the given equation can be rewritten as \(\ln((2x+1)(x-3)/x^2)=0\)

Step 2: Express the logarithm as an exponent

For any \(b^y=x\), it can be rewritten as \(\ln_b(x)=y\). Therefore, you can rewrite the equation from step 1 as \(((2x+1)(x-3)/x^2)=e^0\)

Step 3: Simplify the equation

As \(e^0=1\), the equation from step 2 simplifies to \(((2x+1)(x-3))/x^2=1\). Multiply both sides by \(x^2\) to get \((2x+1)(x-3)=x^2\)

Step 4: Continue simplifying

Expand the left-hand side to get \(2x^2+2x-3x-3=x^2\). Simplify further to \(x^2+2x-3=0\)

Step 5: Solve for x

You can solve for x by factoring, completing the square or using the quadratic formula. After factoring, the equation becomes \((x+3)(x-1)=0\). Setting each factor equal to zero gives two potential solutions: \(x=-3\) or \(x=1\)

Step 6: Check the solutions

Substitute the solutions into the original equation to check if they are valid solutions. \(x=-3\) makes the argument of the second logarithm negative, which is not possible. So, \(x=-3\) is not a solution. However, \(x=1\) is a valid solution.