In Exercises 74–79, solve each logarithmic equation. log2 (x+3) +...

Question

In Exercises 74–79, solve each logarithmic equation. log2 (x+3) + log2 (x-3) =4

Accepted Answer

Hello. Today we're going to be solving a logarithmic equation. The equation given to us is log base three of x plus four Plus Log Base three Of X -4 equal to two. Now before we start this let's just go ahead and recall our multiplication rule for logarithms And that states that if we have two logarithms with the same base A and different arguments and they're being added with each other, then we can combine this into a single logarithms with the same base. But now both arguments are a product of each other. So we're gonna need to do this to the left hand side of this equation. So we have log base three of X plus four plus log base three of X minus four. Combining these together is going to give me log base three of X plus four Times X -4 equal to two. And now we need to figure out a way to get rid of this logarithms. Well there is another rule that states that if we have the logo rhythm of a equal to some arbitrary constancy, we can rewrite this equation as a race of the power of C. Equal to be. So all we're doing is taking this base of A. And raising it to the power of C. And then sending it equal to B. So that's what we're gonna do to this equation. So what we need to do is take the base of three and set it to the power of two and have equal to everything in the argument. Doing so is going to give me three squared equal to X plus four times x minus four. Okay, now we just need to go ahead and simplify in order to open up these parentheses, we're going to need a foil both of these minnow meals together. Doing so is going to give me X squared minus four X plus four, X minus 16 equal to three squared which is nine. Now we have negative four X and positive four X. Those are just going to cancel each other out and now we are left with X squared minus 16 equal to nine. Now again we're solving for X. So we're gonna go ahead and add 16 to both sides of this equation. Doing so is going to give me x squared equal 2 25 And now we want to get rid of this exponent of two in order to do so we're going to take the square root of both sides of this equation. That then leaves us with X equal to plus or - because taking the square root of a number produces a positive and a negative value. Okay, so we have two values for X. However, keep in mind that we are not allowed to have a negative number inside of a logarithms. So we're not going to include the negative value for X. Thus X is going to equal to five and that is going to be our solution. That also means that the solution to this problem is going to be a So I hope this video helps you in understanding how to solve a logarithmic equation, and I'll go ahead and see you all in the next video.

In Exercises 74–79, solve each logarithmic equation. log2 (x+3) + log2 (x-3) =4

Key Concepts

Logarithmic Properties

Exponential Form

Domain of Logarithmic Functions

Watch next