# Solving exponential equations using logarithms: base-2

CCSS Math: HSF.LE.A.4

## Video transcript

Voiceover:Let's say that we've got the function y is equal to five times two to the t power. Someone were to come up to you and say, "Hey look, this is an interesting function." "But I'm curious, I like the number 1,111." "I'm curious at what point for what t value" "will my y be equal to 1,111." I encourage you to pause this video and think about it on your own. At what t value will this y be equal to a roughly equal to 1,111. If you see the need you might want to use a calculator. I'm assuming you've given a go at it. Let's work through this together, so we want to say, when does five times two to the t power, equal 1,111. Let's write that down. When does five times two to the t power equal 1,111. Whenever we're doing anything algebraically it's always a little bit useful to see if we can isolate the variable that we're trying to solve for, we're trying to find what t value will make this equal that right over there. A good first step would maybe try to get this five out of the left hand side, so let's divide the left by five. If we want to keep this being in equality we have to do the same thing to both sides. We get two to the t power is equal to 1,111 over five. How do we solve for t here? What function is essentially the inverse of the exponential function? Well it would be the logarithm. If we say that a to the b power is equal to c then that means that log base a of c is equal to b. a to the b power is equal to c. Log base a of c says what power do I need to raise a to, to get to c? Well I need to raise a to the b power, to get to c. a to the b power is equal to c. These two are actually equivalent statements. Let's take log base two of both sides of this equation. On the left hand side you have log base two of two to the t power. On the right hand side, you have log base two of 1,111 over five. Why is this useful right over here? This is what power do we have to raise two to, to get two to t power? Well to get two to the t power, we have to raise two to the t power. This thing right over here just simplifies to t. That just simplifies to t. On the right hand side, we have log base two, we have all of these business right over here. I'll just write it over, t is equal to log base two of 1,111 over five. This is an expression that gives us our t value but then the next question is well how do we figure out what this is? If you take out your calculator, you'll quickly notice that there is no log base two button, so how do we actually compute it? Here we just have to apply a very useful property of exponents. If we have log base two of well really anything. Let me write it this way, if we have log base a of c, we can compute this as log base anything of c over log base that same anything of a. This anything has to be the same thing. Our calculator is useful because it has a log, you just press log, it's log base 10. If you press ln, it's natural log or log base e. I like to just use the log base 10, so this is going to be the same thing as log base 10 of 1,111 over five over log base 10 of two. We can get our calculator out and we could unlog base e if we wanted that would be a natural log but I'll just use the log button. This is logarithm of 1,111 over five, so that's this part right over here. This is log base 10, implicitly that's what the log button is. Divided by log base 10 of two and then that gives us seven, well it just keeps on going but is approximately equal to 7.796. This is approximately equal to 7.796, so when t is roughly equal to that you're going to have y equaling 1,111.