Video: Solving Logarithmic Equations over the Set of Real Numbers

Solve logβ‚‚ [251 + log₃ (π‘₯ + 7)] = 8, where π‘₯ ∈ ℝ.

02:39

Video Transcript

Solve the equation log base two of 251 plus log base three of π‘₯ plus seven equals eight, where π‘₯ is an element of the real numbers.

Here, we’ve been given a logarithmic equation. This equation involves logs with two different bases. We’ve got log base two over here and log base three over here. Now, we recall that a logarithm is the power to which a number must be raised in order to get some other number.

Let’s take a general logarithm. Let’s say log base 𝑏 of π‘Ž is equal to 𝑐. Now, taking log base 𝑏 is essentially the inverse of raising to the power of 𝑏. So we raise both sides as a power of 𝑏. And we get 𝑏 to the power of log base 𝑏 of π‘Ž equals 𝑏 to the power of 𝑐. And since taking the log base 𝑏 is the inverse of raising it as a power of 𝑏, the left-hand side simply becomes π‘Ž. And so saying log base 𝑏 of π‘Ž is equal to 𝑐 is equivalent to saying that π‘Ž is equal to 𝑏 to the power of 𝑐.

And so we can consider solving our equation in two ways. We could use this general definition, or we could raise both sides as a power of two. When we do so, on the left-hand side, this is the equivalent of the letter π‘Ž in our general form. We’re left with 251 plus log base three of π‘₯ plus seven. And this is equal to two to the eighth power in our general form. That’s 𝑏 to the power of 𝑐. Two to the eighth power is 256. So our equation is 251 plus log base three of π‘₯ plus seven equals 256. Let’s subtract 251 from both sides. And when we do, we see that log base three of π‘₯ plus seven is equal to five.

And now we’re going to perform a similar step to our first step here. We could either use the general definition or equivalently raise both sides. But this time, we do it as a power of three. When we do so, on the left-hand side, we’re left with simply π‘₯ plus seven. And of course, once again, this is equivalent to π‘Ž in the general definition. On the right-hand side, we get three to the fifth power. Three to the fifth power is 243. So our equation becomes π‘₯ plus seven equals 243.

Remember, we’re solving for π‘₯. So, finally, we just subtract seven from both sides. 243 minus seven is 236. And so we’ve solved the equation. We get π‘₯ equals 236. And of course, we could check the solution depending on the functionality of our calculator by substituting π‘₯ is equal to 236 into our original expression and checking we do indeed get eight.

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