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Question Video: Solving Logarithmic Equations Involving Laws of Logarithms and Quadratic Equations Mathematics • 10th Grade

Find the solution set of log₆(−2𝑥 + 48) = 2 log₆ 𝑥 in ℝ.


Video Transcript

Find the solution set of log base six of negative two 𝑥 plus 48 equals two times log base six of 𝑥 in the set of real numbers.

So we write out this equation again. The left-hand side is the logarithm base six of an expression. The right-hand side almost is, but there’s also a factor of two to deal with. Luckily, we have a law of logarithms, which allows us to write the right-hand side as just the log base six of something. This is that 𝑛 times log base 𝑏 of 𝑎 is equal to log base 𝑏 of 𝑎 to the power of 𝑛.

And you can imagine this multiplicative constant 𝑛 moving across to become the exponent of 𝑎. And so on the right-hand side, the multiplicative constant two, which is outside the logarithm, comes inside the logarithm as the exponent of 𝑥. The left-hand side just remains the same. So now on both sides, we have the logarithm base six of some expression. And so the things inside those logarithms on either side must be equal. Negative two 𝑥 plus 48 must be equal to 𝑥 squared.

Another way of saying this would be that six to the power of the left-hand side must be equal to six to the power of the right-hand side. Either way, we get this quadratic equation here and hopefully we know how to solve this. Subtracting negative two 𝑥 plus 48 from both sides, we get that zero is equal to 𝑥 squared plus two 𝑥 minus 48. We can swap both sides to make things conventional so that the zero is on the right-hand side. We can factorize to get 𝑥 plus eight times 𝑥 minus six equals zero. And so the solutions to this quadratic equation are 𝑥 equals negative eight and 𝑥 equals six.

However, because we’re working with a logarithmic equation, we have to be careful that both solutions to the final equation are also solutions to the original equation involving logarithms. The argument of a logarithmic function must be greater than zero. So negative two 𝑥 plus 48 and 𝑥 must both be positive for this original equation to make sense.

Let’s first check 𝑥 equals negative eight. Substituting in negative eight to our original equation, we get log base six of negative two times negative eight plus 48 is equal to two times log base six of negative eight. Simplifying this gives us that log base six of 64 is equal to two times log base six of negative eight.

Negative eight is less than zero of course, and it doesn’t make sense to talk of the logarithm of a negative number. Log base six of negative eight will be the number that you raise six to the power of to get negative eight. And clearly, there is no such number. And so while negative eight is a solution to the quadratic equation that we got, it’s not a solution to the original logarithmic equation that we had to solve.

On the other hand, after substituting in six to our original logarithmic equation and simplifying, we get that log base six of 36 is equal to two times our log base six of six. 36 and six are both greater than zero and so both sides of the equation are defined and meaningful.

And if you think about it or use your calculator, you’ll see that the equation actually is true. Log base six of 36 really does equal two times log base six of six; they’re both equal to two. And so, 𝑥 equals six is a solution to the logarithmic equation. In fact, it’s the only solution. And so the solution set of this equation which is after all what we were asked for is just the set containing the number six.

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