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

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+log(x+3)=log 10

Accepted Answer

Welcome back. I am so glad you're here. We have got this log rhythmic equation. We're told that the log of five X plus the log of X plus two equals the log of 15. And we are to solve for X. We've got a couple of them. They're inside these logs, how do we solve for them? We recall from previous lessons that when we have the log of something plus the log of something, then that can be condensed into the log of those two, something. So in this case five X and X plus two multiplied together. So the log of five X times X plus two and that is still equal to the log of 15. Well this is a lot easier to work with because we recall from previous lessons on the 1-1 property of logarithms that when we have the log of something equal to the log of something else than the something is equal to the something else. Or in this case five X times X plus two is equal to 15 and we can write it just like that and look now there are no logs and now we can solve for X. Let's distribute that five X. So five x times x is five X squared five X times two is plus 10 X equals 15. All right, we've got an X squared on the left. So we've got a quadratic. Let's bring the 15 over. We'll set the right hand side equal to zero. So that possibly we can factor this five X squared plus X minus 15 equals zero. Well, all three of those terms on the left have of five in them. So let's factor out of five. So dividing each term by five gives us five times X squared plus two, X minus three equals zero. Now we can factor that X squared plus two, x minus three. And what times what gives us X squared. That's X times X. What? Two numbers when multiplied, give us a negative three. But when added together give us a positive two. That would be plus three minus one. And now we have three factors five, X plus three and X plus one. And we can set all of the ones with an X in them equal to zero. See what X values would give us a zero there. So X plus three equals zero and x minus one equals zero. And for the first one we can subtract three from both sides. For the second one we can add one to both sides. So X plus three equals zero, solved for X equals negative three. For x minus one equals zero. That solved for x equals one. And now we need to go back to our original equation because for logs we have domain restrictions, what's inside the parentheses needs to be positive. It needs to be greater than zero. So 15 Yes that's positive but five X has to be positive and X plus two has to be positive. So let's plug a negative three in for x. See if it's positive. So this will be five times negative three and this one will be negative three plus two. Well no five times negative three is negative 15 that's negative and negative three plus two is negative one. That's negative, negative three does not work as a solution. Now let's try X equals one. So we've got five times one that's five, that's positive. And we've got one plus two that's three that's positive. So X equals one. Does work as a solution. We look at our answer choices and that matches with answer choice. A well done. We'll catch you on the next one.

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