Problem 84 - Introductory and Intermediate Algebra for College Students 4th

Question

Determine whether each equation is true or false. Where possible, show work to support your conclusion. If the statement is false, make the necessary change(s) to produce a true statement. $$\frac{\log (x+2)}{\log (x-1)}=\log (x+2)-\log (x-1)$$

Step-by-Step Solution

Answer

The original equation is false. The corrected statement is \(\log ((x+2)/(x-1)) = \log ((x+2)/(x-1))\), which is true for \(x>1\).

Step 1: Re-write the Right Side

Re-write the right side of the equation using the quotient rule for logarithms. The quotient rule states that \(\log x - \log y = \log (x/y)\). So, \(\log (x+2) - \log (x-1)\) can be re-written as \(\log ((x+2)/(x-1))\). So the equation becomes \(\frac{\log (x+2)}{\log (x-1)}=\log ((x+2)/(x-1))\).

Step 2: Verify the Equation

At this point, it can be seen that the equation is false. This is due to the structure of the left-side of the equation \(\frac{\log (x+2)}{\log (x-1)}\), which does not follow any of the rules of logarithms, like the one used in step 1.

Step 3: Correct the Equation

To produce a correct statement, we can choose a general way to modify the left side to match the right side. Write the left-side as a single logarithmic expression using the quotient rule. The corrected equation will become \(\log ((x+2)/(x-1)) = \log ((x+2)/(x-1))\). This equation is true for any value of x that satisfies the conditions \(x+2>0\) and \(x-1>0\), i.e. \(x>1\).