Question Video: Using the Inverse Relation between Exponential Functions and Logarithms | Nagwa Question Video: Using the Inverse Relation between Exponential Functions and Logarithms | Nagwa

# Question Video: Using the Inverse Relation between Exponential Functions and Logarithms Mathematics

Solve π₯ + 6 = π^(ln π₯Β²) for π₯, giving your answer to three decimal places.

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### Video Transcript

Solve π₯ plus six is equal to π to the power of ln π₯ squared for π₯, giving your answer to three decimal places.

Well, to start to solve this problem, the first thing we look at is this term here, which is π to the power of ln π₯ squared. To understand what this term will become, what Iβm gonna do is Iβm going to say that π¦ is equal to π to the power of ln π₯ squared. Itβs worth noting that you may also here ln called ln or natural logarithm. But just for this video, Iβm actually gonna say ln.

Okay, so the first thing weβre gonna do is actually take the natural logarithm of each side. So itβs gonna give me that ln π¦ is equal to ln π to the power of ln π₯ squared. And then next, we will apply a log rule that says that ln π to the power of π₯ is equal to π₯ ln π. And this gonna give us that ln π¦ equals ln π₯ squared multiplied by ln π.

And then, weβre gonna use a relationship that we know which is that ln π is equal to one. So therefore, weβre left with ln π¦ is equal to ln π₯ squared because it will be ln π₯ squared multiplied by one, which just gives us ln π₯ squared. And then, we have that π¦ is equal to π₯ squared. And we know that because actually we have a log to the same base of both sides for equation.

Okay, great, so why is this useful? Well, itβs useful because actually we know that π¦ β looking back to the beginning β was equal to π to the power of ln π₯ squared. So therefore, we can say that π to the power of ln π₯ squared is just equal to π₯ squared.

Okay, great, what I can actually do now is look back to our equation and we can actually apply this relationship and substitute π₯ squared for π to the power of ln π₯ squared. So therefore, we have π₯ plus six is equal to π₯ squared. So now, what we do is we actually rearrange to set all equal to zero. So what Iβve actually done is actually subtracted π₯ and six from each side. So we get zero is equal to π₯ squared minus π₯ minus six.

So then, weβre actually gonna solve our quadratic and weβre gonna solve it using factoring. So if we factor it, we get zero is equal to π₯ minus three multiplied by π₯ plus two. And we get this factor because negative three multiplied by two is equal to negative six. And thatβs correct because thatβs what it should equal because it should equal the final term of our quadratic. And then, we have negative three plus two is equal to negative one. And thatβs correct because thatβs actually the coefficient of our π₯ term. So great, theyβre both checked. So yeah, theyβre definitely the correct factors.

Okay, so we can now solve. And in order to find our π₯-values at this point, what we actually want to do is to set each of our parentheses to zero because if the answer of the equation is zero, then at least one of our parentheses will also have to be zero to achieve this. So first of all, we will start with π₯ minus three is equal to zero. So therefore, if we add three to each side, we get that π₯ is equal to three. So great, thatβs one of our solutions.

And then, if we have π₯ plus two is equal to zero, if we subtract two from each side, we will be left with π₯ is equal to negative two. So thatβs our other solution. So therefore, we can say that solutions to π₯ plus six is equal to π to the power of ln π₯ squared for π₯ are π₯ is equal to three or π₯ is equal to negative two.

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