Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 48. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Using Logarithms and Exponentiation to Solve Equations 451 log6 x logA-V Returning to our original problem we convert log35100 to log base 10 as follows. log3 100 _ 1 log 100 _ 1 2 _ 2 5 5 log 3 5 log 3 5 log 3 EXAMPLE SOLUTION EXAMPLE SOLUTION Solve for x if log7 4x 3. In order to solve for x we need to free the x from the logarithm. There are two approaches we can take to this. The flrst is to undo log7 by exponentiating both sides of the equation with base 7. log7 4x 3 7log7 4x 73 4x 73 73 343 x ------ 44 Notice that if we use a base other than 7 to exponentiate then we will get stuck. For instance we could write 10log7 4x 103 but we cannot simplify the expression 10log7 4x so this course of action is unproductive. An alternative mindset for solving the original problem is to think about what log7 4x 3 means and write the statement in exponential form. log7 4x 3 means 3 is the number we must raise 7 to in order to get 4x. So 73 4x .Then x 73 343. Theme and Variation. Examples and are prototypical examples. Example is of the form B x A and Example is of the form log6 x A. Knowing our way around equations of these forms will serve as a guideline for strategizing when we have more complicated examples. Below we show variations on these basic themes. Along the way we ll try to point out some pitfalls so you can walk around them instead of falling into them. It s surprisingly easy either to get caught in a frenzy of unproductive manipulation or to plunge down a short dead-end street when approaching exponential or logarithmic equations. Take time to strategize. Solve for x if 52x 1 20. We need to bring down the exponent to solve for x. Log property iii will help us do this. Again although we can use log6 with any base if we want a numerical approximation then using common logarithms or natural logarithms simplifies the task. 452 CHAPTER 13 Logarithmic Functions 52x 1 20 log 52x 1 log 20 2x 1 log 5 log 20 2x log 5 log 5 log 20 2x log
