Exponential & Logarithmic Equations - Online Tutor, Practice Problems & Exam Prep

Solving Exponential Equations Using Like Bases

In working with exponential functions, we would evaluate our function for some given value of x by simply plugging that value in for x and ending up with an answer. But what if our function is already equal to something? Well, then we're faced with finding the value for x that will then make our statement true. And now that we're faced with solving a new type of equation, you may be worried that we're going to have to learn an entirely new method of solving here, but you don't have to worry about that because I'm going to show you how we'll end up just solving just a basic linear equation that we've solved a million times before by simply doing one thing to our exponential equation and rewriting each side to have the same base. From there, it's literally just solving a basic linear equation.

So let's get right into it. Now here I have the equation 16 is equal to 2 to the power of x. So here I'm looking for the value of x that will make \(2^x\) equal 16. So we want to rewrite these sides to have the same base. And this 2 I know that I can't rewrite as the power of anything, so I'm going to leave that as \(2^x\).

And then I want to rewrite 16 to have that same base of 2. Now I know that 16 is simply equal to \(2^4\), and now both of these sides have that same base of 2. Now, from here, we can simply go ahead and take our powers and set them equal to each other. So setting our powers equal to each other will end up with 4 is equal to x. And I'm actually done here.

I don't even have to solve for anything. But typically, you're going to have to solve for x. Here, since x is already solved for, I already have my answer that x is equal to 4. Now let's go ahead and look at some more examples just to get a bigger picture of what exactly is going on here. So looking at this first example I have, I have 64 is equal to \(2^x\).

Now it might not always be immediately obvious how you can rewrite your powers, so let's break this down a little bit more. Now \(2^x\), I'm not going to rewrite that 2 as anything. It's just going to stay as that. But 64 then needs to get rewritten as a power with base 2. Now I don't know exactly what power 2 needs to be raised to get that, so I can break this down a little bit more.

I know that 64 is equal to \(8^2\), and I also know that 8 is just \(2^3\). So if I take 8, break it down into \(2^3\), and I know that that 8 has to get squared, this now has a base of 2. Now \(2^3\) to the power of 2 is really just \(2^6\). So I have successfully rewritten that side, 64, as a base of 2. Now I know that \(2^6\) is equal to that \(2^x\) that I originally had.

And here my bases are now equal to each other. They're both 2. And I can simply take my powers, 6 and x, and set them equal to each other. So 6 is equal to x. And I don't have anything left to do here.

I already have my answer that x is equal to 6. That will make my statement true. Let's move on to our next example. Here we have \(5^{x+1}\) is equal to the square root of 5. Now both of these already have this 5, so I know that that's what my base needs to be.

And this \(5^{x+1}\) is going to stay \(5^{x+1}\). But I know that the square root of 5 can be rewritten as an exponent. So I can rewrite this as \(5^{1/2}\). Now both of my bases are the same here. They're both 5.

So I can go ahead and take those powers and set them equal to each other. So I end up with the linear equation, \(x+1 = 1/2\). Now I can easily solve for x here by simply subtracting 1 from each side. And I end up with \(x = -(1/2)\). So I have my final answer here, and I just solved a basic linear equation by rewriting like bases.

So let's look at our final example here. We have 27 is equal to \(9^x\). Now looking at this on first glance, you may think, okay, since I have \(9^x\), I need to rewrite 27 with a base of 9. But 27 cannot be rewritten with a base of 9, so I need to get a little bit more creative here. Sometimes you're going to have to rewrite both sides in order to get them to have the same base.

So here I know that 9 is equal to \(3^2\). So I can rewrite this side as \(3^{2x}\) multiplying those exponents. Now 27 I know is equal to \(3^3\). So now I have the same base on both sides of 3.

And I can take those powers, 3 and \(2x\), and set them equal to each other. So I end up with 3 is equal to \(2x\). And in order to solve for x here, I can go ahead and divide both sides by 2. And I'm left with my final answer, that \(x = 3/2\). Now I know that if I plug that back into my equation, that would make my statement true.

Now that we know how to solve exponential equations by rewriting each side to have the same base, let's get some more practice.

Solve the exponential equation.

$4^{x+7}=16$

• A. $2$
• B. $5$
• C. $9$
• D. $-5$

Solve the exponential equation.

$100^{x}=10^{x+17}$

• A. $17$
• B. $34$
• C. $8.5$
• D. $0$

Solve the exponential equation.

$81^{x+1}=27^{x+5}$

• A. $-13$
• B. $11$
• C. $14$
• D. $3$

Solving Exponential Equations Using Logs

Hey everyone. We just learned that whenever we have an exponential equation like 16 is equal to 2 to the power of x, we can simply rewrite each side to have the same base and then set our powers equal to each other in order to get our final answer. But what if we're given an exponential equation like 17 is equal to 2 to the power of x? It's not exactly easy to rewrite 17 as a power of 2. And don't worry, you're not going to have to figure out what crazy decimal that 2 needs to be raised to in order to get 17 because here we can simply solve this equation using logs.

So here I'm going to walk you through how we can just take the natural log of each side and use properties of logs that we already know in order to get our final answer. So let's go ahead and get started. Now let's just walk through an example together and just go through the steps that we're going to need to follow in order to get our answer here for each of these exponential equations. So for our first equation, we have 10^x + 64 is equal to 100. So starting with step 1, the very first thing we want to do is going to be to isolate our exponential expression.

So here our exponential expression is this 10^x, so I want that by itself on one side of my equation. So in order to do that, I need to go ahead and move this 64 over, which I can do by subtracting 64 from each side. Now when I do that I'm going to end up with 10^x is equal to 36. Now from here step 1 is done and I can go ahead and move on to step 2. Now step 2 is when we're going to determine what log we should take.

We're either going to take the natural log or the common log, and here is how we choose. If it has a base of 10, if our exponential expression has a base of 10, which here I have 10^x, I'm going to take the common log, just log of both sides. Now if instead it did not have a base of 10, I would instead take the natural log. These are the only 2 logs you'll use whenever solving exponential equations. So here, like I said, we have a base of 10 so I can go ahead and just take the common log of both sides.

So doing that, I get log of 10^x and that's equal to log of 36. Remember, whatever you do to one side, you have to do to the other. So we are, of course, taking the log of both sides. Now step 2 is done. And we can go ahead and move on to step 3, where we're going to use our log rules in order to get x out of our exponent.

So here, x is in my exponent. But I know that using my power rule, I can go ahead and move that over. So moving that x to the front here, I end up with x times log of 10. And that's equal to log of 36. Now step 3 is done.

We have x out of our exponent. And now we can move on to step 4 and go ahead and solve for x. Now looking at this equation, I have x times log of 10 is equal to log of 36. And you may notice here that I have this log of 10. And since log is just log base 10, I know that this is just going to end up being 1.

So this is really just x times 1, which is just x. So here I end up with x, and that's equal to log of 36. Now here we could be done. X is equal to the log of 36. This is an acceptable answer if you're not asked to approximate because the log or natural log of some number is a constant.

So we really don't need to do anything else from here unless we're explicitly asked to. So step 4 is done. But if you're asked to approximate, we can go ahead and plug this into our calculator. So I would just type in log of 36. And I could end up with a final answer of 1.56.

And I'd be done. Now let's move on to our next example. We have 3 is equal to 2 to the power of x+1. Now we want to restart our steps starting from step 1. And we want to go ahead and isolate our exponential expression.

Now our exponential expression is actually already by itself here, so I don't need to do anything else. Step 1 is done. I can move on to step 2. Now in step 2, we're either going to take the log or natural log. And here, I do not have a base of 10, so I'm going to go ahead and take the natural log of both sides.

So here, I take natural log of 3. And that's equal to the natural log of 2 to the power of x plus 1. Now step 2 is done. And I can go ahead and move on to step 3 and use my log rules to take x out of my exponent. Now again, here, I'm going to use the power rule.

So I'm going to take this exponent of x+1 and pull it to the front of that natural log. So on this left side, I still have the natural log of 3, but now my right side has become x+1 times the natural log of 2. Make sure you pull your entire exponent when using the power rule. So from here, I've taken x out of my exponent. Step 3 is done.

Moving on to step 4, we want to go ahead and solve for x. So this looks a little bit crazy here, but we can simplify this significantly and get x all by itself. So here I have this natural log of 3, and that's equal to x plus 1 times the natural log of 2. But remember that the log or natural log of something is just a constant, and I can treat it just like I would any other number. So I can go ahead and move this natural log to the other side by simply dividing by it like I would any constant.

So dividing both sides by the natural log of 2, I end up with the natural log of 3 divided by the natural log of 2. And that's equal to whatever I have left over here, which happens to be x plus 1. Now I have one final step here in isolating x, getting it by itself, and that is subtracting 1 from both sides. So I am left with that canceling. And I have my final answer here that the natural log of 3 divided by the natural log of 2 minus 1 is equal to x.

Now this looks a little bit scary, but remember, these natural logs are just numbers. So this is really just a bunch of constants, natural log of 3, natural log of 2, and 1. So this is an acceptable answer. But, of course, if you're asked to approximate using a calculator, you can go ahead and plug this into your calculator to get a decimal. Now whenever you do that, you're going to get a final answer of x is equal to 0.58.

And we are completely done here. We have our final answer. Now that we know how to solve exponential equations using logs as well as using like bases, we can solve any exponential equation that gets thrown at us. Thanks for watching, and let me know if you have any questions.

Solve the exponential equation.

$2\cdot10^{3x}=5000$

• A. $x=3.40$
• B. $x=10.19$
• C. $x=0.0001$
• D. $x=1.13$

Solve the exponential equation.

$900=10^{x+17}$

• A. $x=-14.05$
• B. $x=2.95$
• C. $x=0.17$
• D. $x=1.72$

Solve the exponential equation.

$e^{2x+5}=8$

• A. $x=-1.46$
• B. $x=-1.11$
• C. $x=-0.22$
• D. $x=1.39$

Solve the exponential equation.

$7^{2x^2-8}=1$

• A. $x=0.51$
• B. $x=\pm 2$
• C. $x=\pm 2.83$
• D. $x=2.23$

Solving Logarithmic Equations

Hey, everyone. We just solved a bunch of different exponential equations. But what about logarithmic equations? Well, now that we're faced with solving yet another new type of equation, you may be worried yet again that we're going to have to learn something brand new in order to solve these. But, again, you don't have to worry about that at all because here I'm going to show you that there are only ever going to be 2 types of log equations that you run into, 2 logs of the same base that are set equal to each other or a single log set equal to a constant.

And both of them boil down to solving a basic linear equation like we've done a million times before. So let's go ahead and just jump right in. Now with exponential equations, we saw that whenever we had exponents of the same base, we could simply take those powers and set them equal to each other in order to get our answer. And we can actually do the same exact thing whenever working with log equations. So here we have these log base 2.

And since these have the same exact base, I can simply take what we're taking the log of, so \( x + 1 \) and 5, and go ahead and set those equal to each other. So I'm left with just a basic linear equation that I can then solve to get \( x = 4 \). So whenever we have 2 logs of the same base, we can just take what we're taking the log of and set them equal to each other and solve for \( x \). So let's look at a slightly more complicated example here. We have the natural log of \( x+4 \) minus the natural log of 2, and that's equal to the natural log of 8.

Now here, it might not be immediately obvious that we're going to be able to just set some stuff equal to each other, so let's take a closer look here. I know that these are all natural logs, so they are all logs of base \( e \), that same base. And here I can actually use my log properties in order to condense this a little bit more. So I can actually go ahead and use my quotient rule here since I have this subtraction happening. So I can go ahead and condense this into the natural log of \( (x+4)/2 \), and that's equal to the natural log of 8.

Now that these have the same base, I have this natural log on both sides, I can go ahead and set \( (x+4)/2 = 8 \) in order to get the linear equation that I can go ahead and solve for \( x \) with. So this becomes \( x + 4 = 8 \times 2 \) which is 16. Now one final step in isolating \( x \) here is going to be to subtract 4 from both sides, leaving me with \( x = 12 \).

Now that's my final answer, and I'm done. All I had to do was make sure that both sides had the same base and then go ahead and set things equal to each other, giving me a basic linear equation. Now let's look at our other type of log equation. If I cannot rewrite it with like bases and I actually just have a single log, like, say, log base 2 of \( 4x \), and that's equal to a constant. So here you may be worried that this is where it's going to be complicated, but it's not because we're simply going to put this in exponential form and then solve from there.

So let's go ahead and work through this example together. So we have log base 2 of \( 4x \), and that's equal to 5. Now our very first step here is going to be to isolate our log expression which here is log base 2 of \( 4x \). So it's already isolated and I'm already done with step 1. I can move on to step 2 which is the meat of this problem, putting it in exponential form.

Now remember when putting things in exponential form, we're always going to start with that base. So here we're going to start with that 2 and then raise that 2 to the power of 5 and then come back to the other side of our equal sign and set that equal to \( 4x \). So we've completed step 2. This is in its exponential form. We can go ahead and move on to step 3 and actually solve for \( x \). So here I can go ahead and simplify this exponent, this \( 2^5 \). I know that if I multiply 2 by itself 5 times, I'm going to end up with 32. So this is 32, and that's equal to \( 4x \). Now here, to isolate \( x \), I can just divide both sides by 4, canceling on that side, leaving me with my answer that \( x = 8 \).

Now here I already have my answer. I've completed step 3, but we actually have to perform one final check here because we actually want to go ahead to our very last step and check our solution by plugging \( x \) into \( m \). Now \( m \) is just what we are taking the log of. So in this case, it's going to be this \( 4x \) here. So we want to plug our answer into that \( 4x \). And in this case, since we got an answer of 8, we're going to plug in 8 for \( x \). So I'm going to get \( 4 \times 8 \), which is simply 32. And the thing we're looking for here is the sign of this number, whether it's positive or negative. So here, if \( m \) is greater than 0, if it's positive, we are done and this is our solution. So in this case, our solution is good and I know that \( x = 8 \).

But if I got a negative number here, this would actually not be a solution at all. So remember here you cannot take the log of a negative number. It is not a solution if you do get a negative number that you're taking the log of. So now that we know how to solve all of these log equations and we already knew how to solve all of our exponential equations, we are good to go. And let's get some more practice. Let me know if you have questions.

Solve the logarithmic equation.

$\log_3\left(3x+9\right)=\log_35+\log_312$

• A. $20$
• B. $17$
• C. $1$
• D. No Solution

Solve the logarithmic equation.

$\log\left(x+2\right)+\log2=3$

• A. $498$
• B. $1998$
• C. $6$
• D. No Solution

Solve the logarithmic equation.

$\log_7\left(6x+13\right)=2$

• A. $3$
• B. $19.17$
• C. $6$
• D. No Solution