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Solving Equations with Logarithms

Now, let's wrap up this entire unit and our entire course by trying to solve some equations with logarithms. For this equation, we'll have log base 2 of 64 equals x. What I want you to do is to try and change this log form into exponential form, and then solve for x. You can write the value of x here.

Solving Equations with Logarithms

To write this in exponential form we'll have a base of 2, raised to the x power equal to 64. In order to solve for x, we need to rewrite 64 with a base of 2. bases are equal, we know the exponents must be equal as well. So x equals positive 6.

Logarithm Equations Practice 1

What about this equation? What do you think would be the value for x?

Logarithm Equations Practice 1

For this equation x equals positive 3. Nice work if you found 3. We'll start by changing this equation into exponential form. We'll have a base of 3, an exponent of x set equal to positive 27. Next we'll rewrite 27 as 3 cubed, so that way our two bases will match. This must mean that x has to equal positive

Logarithm Equations Practice 2

Let's see if you can solve this equation. What do you think is the value for p, enter it here.

Logarithm Equations Practice 2

P would equal positive 25, in this case. Great work if you found this number. Again, we'll start by converting logarithmic form into exponential form. So we'll have 5 as our base, 2 as the exponent, set equal to p. Remember when we have log of base 5 of p equal to 2. We know that 5 raised to the second power will equal p. So we really just need to evaluate 5 squared. 5 squared is 25, so p equals 25.

Logarithm Equations Practice 3

Try solving this third problem. What do you think would be the value of m? Notice that m is the base here, and also we have a fraction. We have log with base m of 1 16th equal to negative 4. When you think you know the value of m, write it here.

Logarithm Equations Practice 3

Now, this problem is more challenging. But we'll start it out just the same as before. We'll change logarithmic form into exponential form. We'll have m as our base, negative four as the exponent, and 1 16th as the result. So we know m to the negative 4 equals 1 16th. Since we're looking for this base m, we want this to look like something raised to the negative 4th power. If the powers match, then we know what m is. Well, we can rewrite 1 16th as 16 to the negative 1st power. Remember, a negative exponent makes a reciprocal. So 16 to the negative 1 equals 1 16th. Now we can rewrite 16 as 2 to the 4th. We'll have simplify this line. We know that when we raise a power to another power, we really multiply them together. So, 2 to the 4th raised to the negative 1 equals we know that the bases must also be equal for these numbers to both be equal to one another. This is how we get a value for m equal to positive 2. Nice solving, if you got this correct. I know it was tough.

Logarithm Equations Practice 4

Here's another problem where you're looking for the base, what do you think it is?

Logarithm Equations Practice 4

Here, the only solution is positive 5. Nice work if you got this answer. We'll start solving this problem like before. We'll change our equation into exponential form, having the x as our base, 2 as the exponent and set equal to equation. When we take the square root of both sides, we must include the plus or minus symbol for the positive and negative root of 25. We know the square root of 25 equals 5, so x can equal positive 5 or negative 5. But remember for logarithms, the base of a logarithm must be great than zero. So, we know that this value x cannot equal negative 5. So the only value that would make this equation true would be positive 5.

Equivalent Expressions Powers and Logarithms

Now, we've already seen these in practice, but there are some helpful facts that we can use to solve logarithmic equations. The first is that if a to the x equals a to the y, then we know that x must equal y. We saw this earlier in the problem 3 to the x equals 27. We rewrote 27 as 3 cubed. And since we have 3 to the x equal to 3 to the y, we can set this exponent equal to this one. The next fact that can help us solve equations with logarithms is that if a to the x equals b to the x, then we know that the base a must equal the base b. We see this in a problem like this. X squared equals 5 squared. Well we know since these are both squares then x must be equal to 5. In order for this amount on the left, to be equal to this amount on the right. The last fact that we'll use is that if log of base b of n. Equals log base b of n, then we know that m must equal n. Since we're taking the log of base b, we know that these two numbers must be the same in order for the logarithms of those numbers to both be the same as well. We haven't seen an example like this yet, but we'll use this idea in the upcoming problems.

Logarithm Equations Practice 5

Try using that last property that we just covered to solve this equation. What do you think the would be the value for x?

Logarithm Equations Practice 5

In this case, x would equal 4. Great solving if you found this number. Since we have the log of 3 on both sides of the equation, we know that this number must equal this number. So, we'll have 2x minus 1 equal to x plus 3. We'll subtract x from both sides to get x minus 1 equals 3. Then we'll add 1 to both sides to get x equals positive 4. If we use some quick mental math, we can see that this number will be seven and this number will also be seven. So we know the log of base 3 of seven equals the log of base 3 of seven. Now, I don't actually know the log of base 3 of 7 but since the left side equals the right side, I have an equivalent relationship. My equation holds true when x equals 4.

Logarithm Equations Practice 6

Try solving this equation for x. Enter that value here. If there is no solution, then type ns.

Logarithm Equations Practice 6

Again since we have the logarithm of one amount on one side, and the logarithm with base 3 of another amount on the other side. We can set these amounts equal to one another. Now this property only holds true if the bases are the same, and we have one log on the left, and one log on the right. This means we have three x minus 2 equal to x minus 4. Solving for x, I'll subtract x from both sides to get 2 x minus 2 equals negative 4. We'll add 2 to both sides in order to isolate the 2x. And finally, we'll divide both sides by positive 2 to get x is equal to negative 1. But here's where we want to be careful. If we plug in a negative 1 in for x, we'll have log base 3 of negative 5. The same would be true on the right-hand side. If I plug in a negative 1 here, I'll have log base negative 5? What would that equal? Well, we'd be saying 3 raised to some unknown power, equals negative 5. There's no power that we could raise the base we can't take the log of negative numbers, this number must be greater than zero. This means x equals negative 1 isn't a solution so we have no solution.

Logarithm Equations Practice 7 One Log on Each Side

Now, let's try and make this more difficult. Here's another logarithmic equation. And notice that there's multiple logarithms on the left and multiple logarithms with base six on the right. Now, I know this may look complicated, but we can use our properties of logarithms to rewrite the left-hand side and the right hand side. So, what expressions do you think go here, and here? So that way we can simplify the left hand side, and we can simplify the right hand side. Remember that the name of this thing is just called the argument of the logarithm. It's the number that we take the logarithm off. So what are these two arguments?

Logarithm Equations Practice 7 One Log on Each Side

This argument is x squared minus 2x, and this argument is 15. Great thinking if you got these. We know this argument is x squared minus 2x, since we just multiply x minus 2 times x. We multiply the arguments, when we add two logarithms together that have the same base. So x times x equals x squared and x times negative 2 is negative 2x. For the right hand side we're going to multiple 6 times 5 to get 30 and then we'll divide by 2, since for subtracting a logarithm. So 30 divided by 2 equals 15. Remember when we add logs, we multiple the arguments and when we subtract, we divide by that number. This is how we get a number of 15 here.

Logarithm Equations Practice 7

So we have one logarithm set equal to another logarithm, and they have the same base, this means we can set. These two arguments are numbers equal to one another. So what do you think are the values of x? Maybe there's two. Maybe there's only one that will work for the original equation. Or maybe none of them work and there's no solution. Type your answer in here.

Logarithm Equations Practice 7

Here the only solution is x equals positive 5. Nice solving if you found this number. First we'll take our quadratic equation, and we'll subtract 15 from both sides in order to set it equal to 0. So we'll have x squared minus 2x minus 15 equals 0. Now we've factored this quadratic to give us the two factors x minus 5 times x plus 3. Next, we said each factor equal to 0 and then we solve for the values of x. So x can be positive 5 or x can be negative 3. When considering these values for solutions, we want to think back to the original equation. If we plug in a negative 3, in for x, then this argument would be negative, an this one would be as well. Remember, we can't take the logarithm of a negative number. So that means this solution is out. This is why the only answer is x equals positive 5.

Logarithm Equations Practice 8

Try solving this equation for x. Either two solutions, is there only one solution or is there no solution. Remember, if you think it's no solution, type NS.

Logarithm Equations Practice 8

Here x must equal positive 1. Nice problem solving if you figured this out. We start solving this problem by multiplying x times x plus 2. Since we're adding two logs with the same base, we can just multiply these arguments together. So we'll have log base 4 of x times x, which is x squared, and x times 2, which is plus 2x. The log base 4 of 3 stays on the right-hand side of the equation. Now that we have a logarithm on one side of the equation here and a logarithm of something else here, we can set these two things equal to one another. We want to factor this quadratic, but first we need to subtract 3 from both sides, to set it equal to zero. Now factoring, we get x plus 3 times x minus 1. We set each of these factors equal to 0 so x can equal negative 3 or x can equal positive 1. Now keep in mind that we want these arguments to be positive so that way we can take a log of them. So if x were equal to negative 3, I would have a negative number here. I can't take the log base 4 of negative 3. That's not possible. So this solution is out. This means the only solution is x equals positive 1.

Logarithm Equations Practice 9

Try solving this one for x. If there's two solutions, enter them separated by a comma. If there's one solution, just enter that. Or if there's no solutions, type NS.

Logarithm Equations Practice 9

For this problem, x equals 1 3rd. Excellent problem solving if you found this fraction. Again, we'll start by multiplying these 2 arguments together, since we're adding the logs with the same base. So 2x times 3x equals 6x squared. And right hand side. This means we can set this expression equal to 4. Since we have one logarithm of base 3 on one side and another logarithm of base 3 on the other. We subtract 4 from each side of the equation to get 6x squared plus 10x minus 4 equals 0. Now this equation has even coefficients. So I can just divide through by a positive 2. Now I can fator this quadratic. Next, we use factoring by grouping to get the factors of 3x minus 1 and x plus 2. We can set each of these factors equal to 0 and then solve for x. So x can equal positive 1 3rd or x can equal negative 2. When we look back at the original problem, a value of negative 2 for x would make this argument negative, and this argument negative as well. We can't take a logarithm of a negative number, so this solution is out. Our only solution is, x equals one third.

Logarithm Equations Practice 10

Try solving this last exponential equation. Write your solution for x here.

Logarithm Equations Practice 10

The solution for x is positive three halves. Nice work if you found this answer. I really hope you've gained some steam in solving logarithmic equations. We'll start by multiplying these two arguments together to get the log base 6 of 2x squared, plus 7x, equals log base 6 of 15. We get this argument by multiplying x times 2x, which is 2x squared, and x by positive 7, which is positive 7x. Now we can set these two arguments equal to one another. We'll subtract 15 from both sides, to get our quadratic set equal to zero. Factoring this quadratic, will have the two factors of 2x minus 3, and x plus So, x can equal three halves, or one and a half or x can equal negative 5. We want to be cautious and make sure these answers check in our original solution. Since we have log base 6 of x, we know that x equals negative 5 will not be a solution. We can't take the logarithm of a negative number. This solution doesn't work, so our only answer is x equals three halves.