Question: Evaluate/death stare at \(log_{\frac{1}6}(5)\).
Let's talk about what logarithms mean.
Suppose \(log_{\frac{1}6}(5)=x\).
That's the same thing as \(5=(\frac{1}6)^x\). It's just been simplified.
(Logarithms are the inverse operations of exponents)
We can use a calculator to evaluate logarithms that are in base 10.
(In this case, the base is 1/6)
How can we change this so that it uses just base 10?
We can use something called the change-of-base formula.
Here's what the change of base formula looks like.
\(log_x(n)=\frac{log_y(n)}{log_y(x)}\)
In this case, we'll set the base \(y\) to be 10. (you can set it to whatever you want) \(x\) is going to be 1/6, and \(n\) is 5.
When the base is 10, we don't have to write it, it's like a plus zero or a times one.
\(log_\frac{1}6(5)=\frac{log(5)}{log(\frac{1}6)}\)
Punch this into a calculator to find your answer.
\(\frac{log(5)}{log(\frac{1}6)} \approx \boxed{-0.8982444017}\)
You can always check your answer if you need to, of course.
\((\frac{1}6)^{-0.8982444017} \approx 5\)
