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Question Video: Solving Exponential Equations Using Laws of Logarithms Mathematics • 10th Grade

Use a calculator to find the value of 𝑥 for which 2^(2 − 5𝑥) ÷ 5^(𝑥 + 3) = 2. Give your answer correct to two decimal places.

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Video Transcript

Use a calculator to find the value of 𝑥 for which two to the power of two minus five 𝑥 divided by five to the power of 𝑥 plus three is equal to two. Give your answer correct to two decimal places.

In this question, we’re given an equation involving exponents, and we need to solve this equation for the value of 𝑥. We’re told we’re allowed to use a calculator for this question. We need to give our answer correct to two decimal places.

Since this is an equation which involves exponential functions, this should remind us of using logarithms because, remember, logarithmic functions are the inverse functions of exponential functions. There’re several different ways of doing this; we’ll only go through one of these. First, in our equation, notice two of our terms are two raised to some power. In particular, two is equal to two to the first power. So, we’ll solve this equation by taking the log base two of both sides. In fact, we could use any logarithm base; however, using base two will make this equation the simplest.

So we start by taking the log base two of both sides of the equation. This gives us the log base two of two to the power of two minus five 𝑥 divided by five to the power of 𝑥 plus three is equal to the log base two of two. And although we could evaluate the log base two of two by using a calculator, this is not necessary. Remember, we define logarithmic functions to be the inverse of their corresponding exponential functions. In other words, because 𝑎 to the power of one is equal to 𝑎. We must have the log base 𝑎 of 𝑎 is equal to one, where 𝑎 is a positive number not equal to one.

So, we can simplify the right-hand side of this equation. The log base two of two is just equal to one. We now want to simplify the left-hand side of this equation. And to do this, we need to notice we’re taking the logarithm of a quotient. This means we can simplify this by using the following rule, which is often called the quotient rule for logarithms. The logarithm of a quotient is equal to the difference of the logarithms. In other words, the log of 𝑎 over 𝑏 is equal to the log of 𝑎 minus the log of 𝑏.

And it’s worth pointing out here, although we’ve written this as the log base 10, this is actually true for any logarithm base. So, we can apply this to the left-hand side of our equation. Instead of taking the logarithm of this quotient, we’ll take the difference of the logarithms. Doing this, we get the log base two of two to the power of two minus five 𝑥 minus the log base two of five to the power of 𝑥 plus three is equal to one. But we still can’t solve this for our value of 𝑥. Instead, we now need to notice we’re now taking logarithms of power functions. And we know something which will help us simplify the logarithm of a power function.

We know the log base 𝑎 of 𝑥 to the power of 𝑛 is equal to 𝑛 times the log base 𝑎 of 𝑥. This is often called the power rule for logarithms. And all this says is whenever we’re taking the logarithm of a power function, we can instead multiply it by our exponent. Let’s now apply this to both of our terms.

In our first term, we’re now instead going to multiply it by our exponent of two. Doing this, we get two minus five 𝑥 multiplied by the log base two of two. We’ll now do the same for our second term; we’re instead going to multiply it by our exponent of five. This means we’re now subtracting 𝑥 plus three multiplied by the log base two of five. And remember, this is all equal to one. And now, this is in a form which we can start to solve. However, there is one thing worth noticing: the log base two of two is equal to one. So, in fact, this term just simplified to give us two minus five 𝑥. And there’s a reason for this.

Remember, logarithmic functions and exponential functions are inverses. So whenever we’re taking the log base two of two raised to some power, we’re just going to end up with the power. So in fact, we could have skipped using the power rule for this term altogether. So, we’ll just write our first term as two minus five 𝑥.

Next, remember, we’re solving for 𝑥. Let’s distribute the log base two of five over our parentheses. Distributing the log base two of five and the negative over our parentheses, we get negative 𝑥 times the log base two of five minus three times the log base two of five. And now this is just an equation we need to solve for 𝑥. To do this, we need to collect like terms on both sides of our equation. We’ll do this by adding three times the log base two of five to both sides and subtracting two. This gives us negative five 𝑥 minus 𝑥 times the log base two of five is equal to one plus three times the log base two of five minus two.

And on the right-hand side, we can simplify one minus two is equal to negative one. We want to solve this equation for 𝑥. So on the left-hand side of our equation, we’ll take out the shared factor of 𝑥. Doing this, we get 𝑥 multiplied by negative five minus the log base two of five. Finally, all we need to do is divide both sides of this equation through by negative five minus the log base two of five. Doing this, we get 𝑥 is equal to three times the log base two of five minus one all divided by negative five minus the log base two of five. And if we use our calculator to evaluate this expression and write our answer to two decimal places, we get negative 0.81, which is our final answer.

Therefore, we were able to find the value for 𝑥, which solves the equation two to the power of two minus five 𝑥 divided by five to the power of 𝑥 plus three is equal to two. To two decimal places, this was negative 0.81.

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