Using logarithmic laws

[RuiAce]

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The formula sheet provides you the definition
\[ a^x = b \iff x = \log_a (b) \]
and a nice bunch of laws
\begin{align*}
\log_a (x) + \log_a (y) &= \log_a(xy)\tag{sum/product law}\\
\log_a(x) - \log_a(y) &= \log_a\left( \frac{x}{y}\right)\tag{difference/quotient law}\\
\log_a (x^n) &= n \log_a(x)\tag{power law}\\
\log_a(x) &= \frac{\log_b(x)}{\log_b(a)}\tag{change of base law}
\end{align*}
that you don't have to memorise.

But you still need to know how to use them effectively. We'll look at three classic "weird" examples here:

Example 1: Solve \( \boxed{7^{2x-1} = 150}\) correct to 2 decimal places.
There are two common ways of going about these problems. One is to use the definition and jump directly to
\[ 2x-1 = \log_7 (150). \]
But depending on whether or not your calculator can handle base-7 logarithms by default we may need extra work. Your calculator would at least have the base \(e\) log and/or the base \(10\) log built in. So we can use a change of base law!

If I use the base \(e\) logarithm I'll then get:
\begin{align*}
2x-1 &= \frac{\ln 150}{\ln 7}\\
2x &= \frac{\ln 150}{\ln 7}+1\\
x &=\frac12 \left( \frac{\ln 150}{\ln 7}+1\right)\\
&\approx 1.79
\end{align*}
The other approach is to introduce a logarithm to a different base early on. We can take the log of a convenient base on both sides of the initial equation instead. (So I'll use base \(e\) again, because I just have to write 'ln' which is shorter than 'log 10')
\[ \ln 7^{2x-1} = \ln 150 \]
The idea is that now the power law can be used to bring the \(2x-1\) bit down. Eventually the two approaches combine to the same thing!
\begin{align*}
(2x-1)\ln 7 &= \ln 150\\
2x-1 &= \frac{\ln 150}{\ln 7}
\end{align*}
and we do the same thing as above. Some people find this approach to make more sense intuitively and hence they prefer it. Others prefer the first approach just because it's faster.

Example 2: Let \(a = \log_3 2 \) and \(b = \log_3 5 \). Express \( \boxed{\log_3 \frac{75}{4}} \) in terms of \(a\) and \(b\).
Usually, the number that we're taking logarithm of can be somehow decomposed using the first three laws. An easy starting point is to check that firstly the fraction is simplified (which, \( \frac{75}{4}\) is obviously simplified) and then use the difference law immediately.
\[ \log_3 \frac{75}{4} = \log_3 75 - \log_3 4. \]
We'll look at the 'nicer' one first - \(\log_3 4\) looks nicer just because the number is smaller. The idea is to recognise that \(4 = 2^2\), and we're given an expression for \( \log_3 2\) - we assume this equals \(a\)! The power law helps us decompose it a bit:
\begin{align*}
\log_3 4 &= \log_3 2^2\\
&= 2\log_3 2\\
&= 2a
\end{align*}
Now for the uglier one. Bits of intuition are required here, and you'll probably develop them as you go. One possibility is to break 75 into its prime factorisation: \( 75 = 3 \times 5^2\). And the instance we see the product, we realise that the sum law can be used. (And then you'll see the power law used.)
\begin{align*}
\log_3 75 &= \log_3 (3\times 5^2)\\
&= \log_3 3 + \log_3 5^2\\
&= \log_3 3 + 2\log_3 5\\
&= \log_3 3 + 2b.
\end{align*}
So the last piece of the puzzle is to compute \(\log_3 3\). But remember, according to the definition, \(x = \log_3 3\) means that \(3^x = 3\)! Clearly, this would mean \(x=1\), so \(1 = \log_3 3\).

Putting everything back, we therefore have
\[ \log_3 75 = 1 + 2b - 2a. \]

Example 3: Differentiate \(\ln \left(x^3+4x+1)\sqrt{3-x^2} \tan x\right) \).
Let's just assume we now also know the calculus result
\[ \frac{d}{dx} \ln [f(x)] = \frac{f^\prime(x)}{f(x)}. \]
There is a huge temptation to just blindly apply this rule, but it would be disgusting... we'd require a product rule on three terms! The trick is to use log laws to decompose the thing down a bit first:
\begin{align*}
\ln \left(x^3+4x+1)\sqrt{3-x^2} \tan x\right) &= \ln (x^3+3x+1) + \ln \sqrt{3-x^2} + \ln (\tan x)\\
&= \ln (x^3+3x+1) + \frac12 \ln (3-x^2) + \ln (\tan x)
\end{align*}
Okay, so the expression is still ugly. But at least plugging into that derivative rule get us out of the product rule catastrophe!
Remember to differentiate term by term when dealing with sums.
\begin{align*}
\frac{d}{dx} \ln \left(x^3+4x+1)\sqrt{3-x^2} \tan x\right) &= \frac{3x^2+4}{x^3+3x+1} + \frac12 \times \frac{-2x}{3-x^2} + \frac{\sec^2 x}{\tan x}
\end{align*}
You can try simplifying that yourself