# Question Video: Solving Logarithmic Equations Mathematics • 10th Grade

Solve for 𝑥, log₄₉ 𝑥 = −1/2.

03:03

### Video Transcript

Solve for 𝑥, log 49 𝑥 is equal to negative one-half.

This is our log, log base 49 of 𝑥 is equal to negative one-half. The first thing we can do is try to get this side of the equation equal to one. If we multiply both sides of the equation by negative two, we’ll end up with negative two times the log base 49 of 𝑥 equals one. Negative two times negative one-half equals positive one. At this point, we’ll want to remember one of our log rules with the form 𝑘 times log base 𝑏 of 𝑚, where 𝑘 is a constant. We can rewrite that. It’s equal to the log base 𝑏 of 𝑚 to the 𝑘 power.

In our example, the constant 𝑘 is negative two. So, we’ll rewrite this to say log base 49 of 𝑥 to the negative two power is equal to one. At this point, we’ll need to consider another log rule, that log base 𝑏 of 𝑏 is equal to one. Since we know that this log equals one, we can say that 𝑥 to the negative two power is equal to 49. 49 equals 𝑥 to the negative two power. We could rewrite that to say 49 equals one over 𝑥 squared.

If 49 equals one over 𝑥 squared, we can multiply both sides of the equation by 𝑥 squared. And say that 𝑥 squared times 49 equals one. Then, we divide both sides by 49, and we see that 𝑥 squared equals one over 49. So, we take the square root of both sides of the equation. And 𝑥 will be equal to plus or minus the square root of one over the square root of 49. The square root of one is one. And the square root of 49 is seven. At this point, we have 𝑥 being equal to plus or minus one-seventh.

However, when we go back to look at our original 𝑥, we see that 𝑥 is what we’re taking the log of. And we can’t take the log of a negative number. And so, we say that 𝑥 is not equal to negative one-seventh. It’s only equal to the positive result; 𝑥 equals one-seventh. As a check, we enter into our calculator log base 49 of one-seventh. And we would expect it to give us negative one-half, which it does.