Video: Solving Logarithmic Equations over the Set of Real Numbers

Solve logβ‚… (log₃ (7π‘₯ + 194)) = 1, where π‘₯ ∈ ℝ.

02:45

Video Transcript

Solve the equation log base five of log base three of seven π‘₯ plus 194 equals one, where π‘₯ is an element of the set of real numbers.

Here, we’ve been given a logarithmic equation. Now, we’re going to need to be a little bit careful because we have two different bases in this equation. We’ve got log base five and log base three. And so let’s recall what we actually mean by log base 𝑏 of π‘Ž equals 𝑐.

Logarithm is the inverse function to exponentiation. It’s the power to which a number must be raised in order to get some other number. Now, let’s raise both sides of our general form as a power of 𝑏 such that 𝑏 to the power of log base 𝑏 of π‘Ž equals 𝑏 to the power of 𝑐. Now, since a logarithm is the inverse function to exponentiation, 𝑏 to the power of log base 𝑏 of π‘Ž is simply π‘Ž. And so log base 𝑏 of π‘Ž equals 𝑐 is equivalent to saying that π‘Ž is equal to 𝑏 to the power of 𝑐.

So let’s go back to the equation in our question. We’ve got log base five of some other algebraic expression. So let’s raise both sides of our equation as a power of five. When we do so on the left-hand side, we’re simply left with log base three of seven π‘₯ plus 194. In our general form, this is the equivalent to π‘Ž. Raising the right-hand side as a power of five, and when we get five to the power of one. Now, in our general form, that’s 𝑏 to the power of 𝑐. Now, of course, five to the power of one is just five. So our equation becomes log base three of seven π‘₯ plus 194 equals five.

Now, we have a logarithmic equation purely base three. So to solve, we’re going to raise both sides as a power of three. This time, when we do this, we’re left with seven π‘₯ plus 194. And again, this corresponds to the letter π‘Ž in our general form. On the right-hand side, we get three to the fifth power. Now, three to the fifth power is 243. So we’re actually left with quite a simple equation in π‘₯. It’s seven π‘₯ plus 194 equals 243.

Let’s solve this equation by subtracting 194 from both sides. That gives us seven π‘₯ equals 49. Finally, we divide through by seven, and we get π‘₯ is equal to seven. And so, we’ve solved the equation log base five of log base three of seven π‘₯ plus 194 equals one. π‘₯ is equal to seven. And, of course, it’s useful to recall that we can check our answer as we would check the solution to any other equation by substituting π‘₯ equals seven back into the original expression. And we should indeed get one.

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