Solve each logarithmic equation in Exercises 49–92. Be sure to...

Question

Solve each logarithmic equation in Exercises 49–92. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. log(x+4)−log 2=log(5x+1)

Accepted Answer

Welcome back. I am so glad you're here. We are given this log rhythmic equation. We're told that the log of 12 X plus seven minus the log of nine equals the log of two X plus one. And we are asked to solve for X. We've got a couple of excess here and there inside these logs. How do we get them out while we recall from previous lessons on expanding and condensing logs that when we have the log of something minus the log of something that the two some things in here get divided by each other. So you rewrite this as the log of and the first one goes in the numerator. So 12 X plus seven divided by the second one goes in the denominator divided by nine and that is equal to the log of two X plus one still. Well this is actually much more helpful because we recall from previous lessons on the 1 to 1 property of logarithms that when we have the log of something equal to the log of something else, that something and something else are equal to each other. So we can rewrite this as 12 X plus 7/ equals two X plus one and look at that, the logs are gone, we can solve for X. Yay, alright, let's start off by multiplying both sides by nine. So on the left the nines cancel out and we've got 12 X plus seven and on the right will distribute that nine. So nine times two X is 18 X and nine times one is plus nine. Let's start off by moving the excess over to the right hand side of the equation and let's move the constants over to the left hand side of the equation. So subtracting 12 x from both sides on the right, that will leave us with six x and subtracting nine from both sides. 7 -9 is a -2. Next we will divide both sides by six and we have X equals a negative 2/6 which is simplified as a negative one third and written as a decimal. That's negative 0.3 repeating to two decimal or three decimal places. Excuse me, that's negative 0.333. Well great we've got a solution but with the logs we always need to go back to our original equation and make sure that we are within our domain. Our domain restriction for logs is that what's inside the parentheses here? The log of needs to be a positive number. It needs to be greater than zero. So we need to ensure that with our solution of X equals negative one third. All of these log values are positive. So we go back in and plug in a negative one third. So 12 times negative. One third plus 7, 12 times negative one third is negative four plus seven negative four plus seven is a positive three. Yay that is positive, nine is positive and now we've got two times a negative one third +12 times negative one third is negative, two thirds plus one, negative two thirds plus one is a positive one third. Yay, they are all positive so X equals negative. One third is a solution for this equation. We look at our answer choices and that matches with answer choice. A well done, we'll catch you on the next one.

Key Concepts

Properties of Logarithms

Domain of Logarithmic Functions

Solving Logarithmic Equations

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