Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig ReviewThe first step is to divide by the 4, then we’ll convert to an exponential equation.1log ( 1− 5x)=21log( 1−5 x)210 = 101− 5x= 10Note that since we had a common log in the original equation we were forced to use abase of 10 in the exponential equation. Once we’ve used Property 4 to simplify theequation we’ve got an equation that can be solved.121− 5x= 10− 5x=− 1+10121211 ⎛ ⎞2x =− ⎜− 1+10 ⎟5 ⎝ ⎠x = -0.4324555320Now, with exponential equations we were done at this point, but we’ve got a littlemore work to do in this case. Recall the answer to the domain of a logarithm (theanswer to Problem 9 in the Logarithm Properties section). We can’t take thelogarithm of a negative number or zero.This does not mean that x = − 0.4324555320 can’t be a solution just because it’snegative number! The question we’ve got to ask is this : does this solution produce anegative number (or zero) when we plug it into the logarithms in the originalequation. In other words, is 1− 5x negative or zero if we plug x = − 0.4324555320into it? Clearly, (I hope…) 1− 5x will be positive when we plug x = − 0.4324555320in.Therefore the solution to this is x = − 0.4324555320.Note that it is possible for logarithm equations to have no solutions, so if that shouldhappen don’t get to excited!⎛ x ⎞2. 3 + 2ln ⎜ + 3⎟=−4⎝7⎠SolutionThere’s a little more simplification work to do initially this time, but it’s not too bad.© 2006 Paul Dawkins 95http://tutorial.math.lamar.edu/terms.aspx