Video: Solving Logarithmic Equations by Factorisation

Solve (logβ‚… π‘₯)Β² βˆ’ 6 logβ‚… π‘₯ + 8 = 0, where π‘₯ ∈ ℝ.

02:43

Video Transcript

Solve log base five of π‘₯ squared minus six log base five of π‘₯ plus eight equals zero, where π‘₯ is an element of the real numbers.

Now, the fact that we have two terms in log base five of π‘₯ shouldn’t throw us off here. What we need to spot is we have a special type of quadratic equation. And so what we’re going to do is make a substitution. Now, we don’t necessarily need to do this, but it can make the whole process a little less complicated. We’re going to let 𝑦 be equal to log base five of π‘₯. Then log base five of π‘₯ all squared becomes 𝑦 squared. Six log base five of π‘₯ is six 𝑦. And so our full equation becomes 𝑦 squared minus six 𝑦 plus eight equals zero.

And now, we have quite a straightforward, quadratic equation. We can solve for 𝑦 by first factoring the expression on the left-hand side. It will be the product of two binomials whose first term is 𝑦. And that’s because when we redistribute these, we get 𝑦 times 𝑦 which gives us the 𝑦 squared. And so next, we’re looking for two terms whose product is eight and whose sum is negative six. Well, those two numbers are negative four and negative two. Remember, negative four times negative two is positive eight. But their sum is negative six as required.

So our equation is 𝑦 minus four times 𝑦 minus two equals zero. Well, 𝑦 minus four and 𝑦 minus two are just numbers. And the only way for the product of these two numbers to be equal to zero is if either 𝑦 minus four is equal to zero or 𝑦 minus two is equal to zero. And so to solve the first equation, we add four to both sides. And we find 𝑦 is equal to four. Similarly, for our second equation, we add two to both sides. And we get 𝑦 is equal to two. But we’re not finished. Our original equation is in terms of π‘₯. And we’re looking to solve for π‘₯. So we go back to our earlier substitution. We let 𝑦 be equal to log base five of π‘₯. And so we replace 𝑦 with log base five of π‘₯. And we find our two equations are now log base five of π‘₯ is equal to four and log base five of π‘₯ is equal to two.

So how do we solve these? Well, essentially, the logarithm answers the question for us. It answers what exponent do we need for one number to become another number. Here, five is the base. Four is the power or the exponent. And then π‘₯ is the number that becomes. So we can actually rewrite this. And we can say that π‘₯ must be equal to five to the fourth power, which is 625. Similarly, in our second equation, five is the base, two is the exponent, and, this time, π‘₯ is the number that becomes. So π‘₯ is equal to five to the power of two or five squared, which is equal to 25. And so the solution set to the equation log base five of π‘₯ squared minus six log base five of π‘₯ plus eight equals zero, where π‘₯ is an element of the real number set, is 25 and 625.

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