Video: Evaluating Logarithms

What is the value of log_(√5) (1/(125 √5))?

03:54

Video Transcript

What is the value of log base square root of five of one over 125 times the square root of five?

We want to try to rearrange this equation so we can simplify and solve. To do that, we’ll need to use a little bit of creativity. The first thing we notice is that we’re dealing with one over 125 times the square root of five. If we want to think about how we can simplify this, remember that we have an exponent rule that says one over π‘₯ is equal to π‘₯ to the negative one power, which means we could rewrite one over 125 times the square root of five as 125 times the square root of five to the negative one power.

This gets us a bit closer because we know that log of π‘₯ to the π‘Ž power is equal to π‘Ž times log of π‘₯. That means we’ll have negative log base square root of five of 125 times the square root of five. At this point, we need to do something with this 125. But what can we do? How can we rewrite it? If we say that 125 equals five times 25 and 25 equals five times five, we can say that 125 equals five cubed. But what would be the most helpful would be to write 25 as the square root of five to some power.

If we think about the value five, we could say that five equals the square root of five times the square root of five. We could say five equals the square root of five squared. And that means 125 would be equal to the square root of five squared to the third power. And if we have π‘₯ to the π‘Ž power to the 𝑏 power, we can simplify that by having π‘₯ to the π‘Ž times 𝑏 power. So, we could write 125 as the square root of five to the sixth power. Again, this is not something that we commonly do. But our goal was to have a base of the square root of five that we’re taking the log of.

This means in our log equation, we can substitute the square root of five to the sixth power in for 125. We then have inside the log the square root of five to the sixth power times the square root of five to the first power. And based on our exponent rules, π‘₯ to the π‘Ž power times π‘₯ to the 𝑏 power equals π‘₯ to the π‘Ž plus 𝑏 power. Six plus one is seven. So, we now have the simplified expression of the negative log with the base square root of five of the square root of five to the seventh power.

And since we have an exponent in our log now, we can go back to the rule we used earlier and say negative seven times log base square root of five of the square root of five. And this is the key we were looking for the whole time because the log base square root of five of the square root of five is asking the square root of five to what power equals the square root of five. And this is where we get the rule that the log base 𝑏 of 𝑏 equals one because the square root of five to the first power equals the square root of five. And that means in place of the log base square root of five of the square root of five, we’ll have one. In our final step, we’ll just multiply negative seven times one and get negative seven.

The value of the log base square root of five of one over 125 times the square root of five is negative seven.

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