In Exercises 74–79, solve each logarithmic equation. log4 (2x+1) ...

Question

In Exercises 74–79, solve each logarithmic equation. log4 (2x+1) = log4 (x-3) + log4 (x+5)

Accepted Answer

Hello. Today we're going to be solving a logarithmic equation. So the equation given to us is log base of two of three, x plus five equal to log base of two of x minus one plus log base of two of X plus four. Now in order to solve this equation we're going to need to reduce the right hand side first. And in order to do that let's go ahead and recall our logarithmic property for multiplication. And that states that if we have two logarithms with the same base a being added to each other, then you can combine this into a single law algorithm with the same base A where both of the arguments are a product of each other. So we're gonna use this rule on the right hand side of this equation. So we're going to combine both of these logarithms into a single logarithms. Doing so is going to give us log base two of three. X plus five equal to log base two of x minus one times X plus four. Okay now we have the logo rhythm of base two on the left hand side and logarithms base two on the right hand side while there's an X. There's a logarithmic rule that states that if you have an equation with the same log base on both sides of the equation you can just set the arguments of those logarithms equal to each other and that's what we're gonna do here. So again both sides of this equation have the same log base so we can just set both of these arguments equal to each other. Doing so is going to give me three X plus five equal to x minus one times X plus four. Okay now we need to go ahead and sell for X and in order to start that let's go ahead and foil both of these quantities together. Doing so is going to give me three X plus five equal to X squared plus four, X minus x minus four. And we're gonna go ahead and combine like terms together and we're gonna receive three X plus five equal to X squared plus three x minus four. Now, since we're solving for X, we want to go ahead and set this equation equal to zero in order to do that, let's just go ahead and subtract three X to both sides of the equation and subtract five to both sides of this equation. Doing so is going to get me zero equal to X squared minus nine. Now let's go ahead and get X by itself in order to do that, let's add nine to both sides of this equation. Doing so is going to give me X squared equal to nine and finally we want to go ahead and get rid of this exponent of two in order to do that. We're going to take the square root of both sides of this equation. That's going to give me X equal to plus or -3. Now keep in mind it's plus or minus because whenever you take the square root of a number it produces a positive and negative value. Okay, so we have two values for X positive three and negative three. Keep in mind that we never want to have a negative number inside a logarithms as that contradicts its domain. So we're not going to include negative three in our solution in our solution. Thus the answer to this problem is going to be X is equal to three. That also means that the solution to this problem is going to be D. 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. log4 (2x+1) = log4 (x-3) + log4 (x+5)

Key Concepts

Logarithmic Properties

Exponential Form

Domain of Logarithmic Functions

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