Video: Solving Exponential Equations

Solve 2^(π‘₯) = 5𝑒^(π‘₯ βˆ’ 1) for π‘₯, giving your answer to three decimal places.

04:14

Video Transcript

Solve two to the power of π‘₯ equals five 𝑒 to the power of π‘₯ minus one for π‘₯, giving your answer to three decimal places.

So, the first thing we’re gonna do to enable us to solve this problem is take log to the base 𝑒 of both sides of the equation. And when we do that, we get log to base 𝑒 of two to the power of π‘₯ equals log to base 𝑒 of five 𝑒 to the power of π‘₯ minus one. But why do we do this? Well, we did this because we know one of our log laws. And that is that if we have log to the base π‘Ž of π‘Ž, it’s equal to one. So therefore, if we have log to base 𝑒 of 𝑒, it’s also gonna be equal to one, which is gonna come very useful later on.

Now, I’ve written log to the base 𝑒 to help us understand what’s happening. But actually, we refer to log to the base 𝑒 as the natural logarithm or ln. And that’s what I’ll use as we continue the question on. Now to solve the problem, what we have are three laws that refer to natural logarithms that are gonna become very useful. The first of these is the product rule. So if we have ln π‘₯ multiplied by 𝑦, this is equal to ln π‘₯ plus ln 𝑦. Then, the second rule is referred to as the quotient rule. And this tells us if we have ln π‘₯ divided by 𝑦, this is equal to ln π‘₯ minus ln 𝑦. And then thirdly, we have a law that we’ve referred to as the power law. And this tells us that if we have ln π‘₯ to the power of 𝑦, this is equal to 𝑦 multiplied by ln π‘₯. So, great, we’ve got these. And now let’s use them to solve the problem.

So, what we’re gonna do first is use the product rule. I’m gonna use this to split up the expression we have on the right-hand side of our equation. So, when we do that, we get ln two to the power of π‘₯ is equal to ln five plus ln 𝑒 to the power of π‘₯ minus one. Then, next, what we’re gonna do is we’re gonna use the third rule, the power rule. And we’re gonna use this to the exponents or powers that we have in our equation. And when we do that, we get π‘₯ ln two is equal to ln five plus π‘₯ minus one ln 𝑒.

Now, at this point, we can refer back to what we mentioned at the beginning. And that is that if we have log to the base π‘Ž of π‘Ž, this is equal to one. Well, in our case, log to the base 𝑒 of 𝑒, or ln 𝑒, this is just gonna be equal to one. So we know that this part of our equation is gonna be equal to one. So now, what we can do is subtract π‘₯ minus one from each side of the equation. And when we do that, we get π‘₯ ln two minus π‘₯ minus one equals ln five. So then, if we distribute negative one across our parentheses, we get π‘₯ ln two minus π‘₯ plus one equals ln five. So then, we can subtract one from each side of the equation. So we get π‘₯ ln two minus π‘₯ equals ln five minus one.

Now taking a look at the left-hand side of the equation, we can see that we’ve got an π‘₯ in both terms. So, we’re gonna take this out as a factor. So therefore, we get π‘₯ multiplied by ln two minus one is equal to ln five minus one. So then, if we divide each side of the equation by ln two minus one, we’re left with π‘₯ is equal to ln five minus one over ln two minus one. So, we could just leave it like this. But if we read the question, it wants the answer to three decimal places. So what we’re gonna do is actually calculate the answer.

And when we do that, we get π‘₯ is equal to negative 1.9860, et cetera. As we already said, I’m going to do it to three decimal places. So I’ve put a line after the third decimal place, looked at the number to the right of it, which is a zero. So therefore, what’s gonna happen is it’s gonna round down. So we’re gonna stay with six to the left of it. So therefore, we can say the solution to three decimal places of two to the power of π‘₯ equals five 𝑒 to the power of π‘₯ minus one is π‘₯ is equal to negative 1.986.

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