Module 6: Lesson 6

 

Self-Check 1

2. Questions 4.a., 4.c., and 6 on page 413 of the textbook

   4.  
      1. Attempt to verify x = 0.
         
         
          
         
         
         

         Because the logarithm of a negative number is undefined, the solution x = 0 is extraneous.

         
         

         Attempt to verify x = 5.
         
         

          

         
         
         Thus, x = 5 is a valid solution.

      3. Attempt to verify x = −2.
         
         
          
         
         
         Thus, x = −2 is a valid solution.
         
         Attempt to verify x = −6.
         
         
          
         
         
         Because the logarithm of a negative number is undefined, the solution x = −6 is extraneous.
   6.  
      1. When Rubina did her work, she neglected to apply the quotient law of logarithms properly on the left side of the equation (she subtracted the terms instead of dividing them). The proper solution to the question is as follows:
         
         
          
         
         
         Verify the solution.
         
         
          
         
         
         Thus, x = 2 is a valid solution.
      2. Ahmed did everything correctly up until the last step. While x = −6 is extraneous, x = 0 is a solution.
         
         

         Verify the solution.
         
         

          

         
         
         Thus, x = 0 is a valid solution.

      3. In the third step, Jennifer eliminated the logarithm on the left side of the equation. This cannot be done because there is not a logarithm with the same base on the right side of the equation. A possible method for correctly solving the equation is shown:
         
         
          
         
         
         Because the value x = −4 would result in negative logarithms if substituted back into the original equation, the value is extraneous. Verify x = 2.
         
         
          
         
         
         Thus, x = 2 is a valid solution.

© 2012 Alberta Education