Logarithms

Published: Dec 27th, 2020

In Lesson #7: Four Basic Functions, we discussed the inverse function. Today, we'll discuss a specific inverse function; the logarithm!

Below is a sketch of \(f(x) = 2^x\) plotted against \(f(x) = \log_2 x\):

As you might notice, they are the same function reflected along a particular line? Care to guess the line's equation? 

Well, as you might remember from Lesson #4, a property of the inverse function is that it is reflected along the line \(y=x\)!

Now, you might be wondering, "how is it useful to know that logarithms are the inverse of exponential functions?" I'll show you.

An exponential function takes some input and raises it as the power of some base. For instance, if my function was \(f(x) = 2^x\) and my input was \(x=1\), then \(f(1) = 2^1 = 2\)

What would you imagine a logarithm would do, then? Well, the inverse of \(2^x\) is \(log_2 x\) (log base two of \(x\)).

If I now use 2 as the input for this log (the output of the exponential function), then what do you think my output will be?

\(f(2) = log_22 = 1\)! We're back where we started! 

Therefore, the logarithm of some base tells us what exponent we should raise the exponential function with the same base to get a given value.

For our function \(f(x) = log_2x\), we're asking "two to the power of what will give \(x\)?" 

Confused yet? I don't blame you. Let's try some examples to clarify:

Solve \(log_464\) for \(x\)

This question is asking us: to what power do I need to raise 4 to get 64?

Well, as you might know, \(4^3 = 64\), so our answer is 3.

Solve \(log_{10}1000\)

This question is asking us: to what power do I need to raise 10 to get 1000?

Well, as you might know, \(10^3 = 1000\), so our answer is 3.

In general, unless there's some other base placed as a subscript, assume that \(logx\) refers to the base 10 log of x.

Another special logarithm type is \(ln\), or the natural logarithm, which refers to \(log_e\), but don't worry about that!

Let's continue, shall we? 

Solve \(5^x = 400\) 

Hmmm. The bases aren't the same here. Luckily, however, logarithms offer a solution!

\(log 5^x = log 400\)

Now comes the cool trick:

\((x)(log 5) = log 400\)

WHAAAAT! That's right, logarithms allow us to bring that exponent down as a coefficient!

\(loga^{x} = x \cdot loga\)

Now, all we need to do is solve for \(x\):

\(x = \frac{log 400}{log5} \approx 1.903\)

Aaaand, there we have it. Logarithms. Now, be sure to try some practice problems!

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