Log Laws and Change-of-Base Formula
Log Laws
Just like how exponential expressions have laws for multiplying them (i.e. adding exponents, like with \(2^3\cdot 2^5=2^8\)), dividing them (i.e. subtracting exponents, like with \(\frac{2^5}{2^3}=2^{2}\)), etc., we have analogous laws for logarithms! And it turns out that the laws for logs largely are consequences of the laws for exponents (because logs, in a sense, “undo” what exponents do).
For \(a>0\), and \(m\), \(n\) either both positive or both negative real numbers,
$$\log_a(m\cdot n)=\log_a(m)+\log_b(n).$$
That is, if you have multiplication on the inside of a log, you can convert it to the SUM of two logs. In other words, the log of a product is a sum of logs!
Seeing the law with numbers instead of variables:
$$\begin{align}\log_2(4x)&=\log_2(4)+\log_2(x)\\&=2+\log_2(x)\end{align}$$
For \(a>0\) and \(m,n\) either both positive or both negative, and \(n\neq 0\) (because we can’t have division by \(0\)),
$$\log_a\left(\frac{m}{n}\right)=\log_a(m)-\log_a(n).$$
That is, when you have division on the inside of a logarithm, you can convert it to a difference of two logs. In other words, the log of a fraction/quotient is a difference of logs.
Seeing the law in action with some numbers:
$$\begin{align}\log_2\left(\frac{8}{3}\right)&=\log_2(8)-\log_2(3)&=.
For \(a,m>0\), and where \(n\) is any real number,
$$\log_a(m^n)=n\cdot \log_a(m).$$
That is, when you have an exponential expression on the inside of a log, you can basically move the exponent out in front of the logarithm, multiplying it to whatever is already there. For example,
$$\begin{align}5\log_2(3^7)&=5\cdot 7 \log_2(3)\end{align}$$
For \(a>0\), and when \(n\) is a real number,
$$\log_a(a^n)=n.$$
This is a special case of the last law. Notice that \(log_2(2^n)=n\cdot\log_2(2)=n\cdot 1=n,\) thus proving this law.
This law states that when the base of the log and the base of an exponential expression inside the log are the same, then the bases cancel.
Quick examples, \(\log_2(2^3)=3\) or \(\log_5(5^{30})=30\).
If \(a>0\) then
$$\log_a(a)=1.$$
This is a special case of the last cancellation law. Notice that \(\log_a(a)=\log_a(a^1)=1\log_a(a)=1\cdot 1=1\).
So, this law states that if the inside of the log and the base match, then they cancel out, giving you just a \(1\).
Quick examples: \(log_2(2)=1\) or \(log_{123}(123)=1\)
If \(a\) and \(m\) are positive real numbers, then
$$a^{\log_a(m)}=m.$$
This law states that if the base of the exponential expression matches the base of the log in the exponent, then the log and the base cancel out, leaving you with what’s inside the log.
Quick examples: \(2^{\log_2(5)}=5\) or \(e^{\ln(x)}=x\)
This law basically states that no matter what base your log is, the log of a negative number or zero DOES NOT EXIST. For instance \(\log_2(-50)\) doesn’t exist. Similarly, \(ln(0)\) also doesn’t exist.
To see why this law is true, consider evaluating the expression $$\log_2(0).$$ This is the same thing as asking “\(2\) to what power gives you \(0\)?”, or in other words, fill in the “?” in the equation below:
$$2^?=0.$$
You will find that you can’t find such a number! The same thing would happen if you replaced \(0\) with a negative number instead.
The Change-of-Base Formula
You have probably wondered at some point over the last few lessons “how can I plug something like \(log_3(2)\) into my calculator.” It is true that most calculators don’t have a generic “log” button that allows you to choose the base and argument (i.e. the thing inside parentheses), so we need a workaround for those of us with more primitive devices.
So basically the change-of-base formula allows us to take any arbitrary log of any base and turn it into a fraction of natural logs. This is good because there IS (very likely) a “\(ln\)” button on your calculator.
Using the change-of-base formula, note that
$$\begin{align}\log_2{3}=\frac{\ln(3)}{\ln(2)}\end{align}$$
Then, plugging each of \(\ln(3)\) and \(\ln(2)\) into your calculator, you get
$$\ln(3)=1.0986123\ \ \ \ and\ \ \ \ \ln(2)=0.693147$$
So we have
$$\begin{align}\log_2{3}=\frac{\ln(3)}{\ln(2)}=1.584963\end{align}$$
Note also that there is nothing special, really, about the use of natural logs in the change-of-base formula. In fact we can use any base we like! This has significance in many algebraic contexts where cancellation of exponential expression bases is called for. We won’t worry too much about that right now though.
Use Log Laws to expand each of the following logarithmic expressions
$$1+\log_2(p)$$
Note
$$\begin{align}\log_2(2\cdot p)&=\log_2(2)+\log_2(p)&\text{Multiplication Law}\\&=1+\log_2(p)&\text{Cancellation Law,}\ \ \log_2(2)=1\end{align}$$
$$3+\log_3(p)-\log_3(q)$$
$$-6\log_2(t)-12\log_2(y)$$
$$\log_2(2y+5x)$$
That is, you can’t do anything!
We don’t have a rule for what to do when you are adding or subtracting stuff inside a log. So we can’t do anything at all because there is no way to turn the inside of the log, \(2y+5x\) into a product or a fraction by factoring. We’re stuck with what we’ve got.
\(\frac{2}{3}\log_5(s)+\frac{4}{3}\log_5(a)-1-\frac{2}{3}\log_5(n)\)
\(\frac{1}{2}\log_3(x)+\frac{1}{5}\log_3(y)-2\log_3(y)-\frac{1}{3}\log_3(t)\)
\(2\ln(p)+\frac{2}{3}\ln(q)-\frac{1}{2}-\frac{1}{5}\ln(x)\)
Use log laws to rewrite the expression so that your final answer has at most one logarithm in it.
\(\log_2(xy)\)
$$\log_3(x^2\cdot z^3)$$
$$\ln\left(\frac{t^2z}{y^3}\right)$$
$$\log_3\left(\frac{t}{y^2}\right$$
$$\log_2((2x+1)(2x-1))\ \ \ \ or\ \ \ \ \log_2(4x^2-1)$$
$$\log_2\left(\frac{x+1}{x-1}\right)$$
\(\log_3(x+3)\).
$$\log_2\left(\frac{(x+3)^2(x+2)^5}{x+1}\right)$$
$$\log_2\left(\frac{(x+3)(x+1)}{(x+2)^{1/3}}\right)$$
$$\log_2(x+1)-\log_3(x+1)+\log_4(x+1)$$
Use the Change of Base Formula to simplify or compute the following expressions.
$$1.584962$$
$$\begin{align}\log_2(3)&=\frac{\ln(3)}{\ln(2)}\\&=\frac{1.098612}{0.6931471}\\&=1.584962\end{align}$$
$$2.153382$$
$$\begin{align} \log_5(32)&=\frac{\ln(32)}{\ln(5)}\\&=\frac{3.4657359}{1.6094379}\\&=2.153382\end{align}$$
$$-4.523562$$
$$\begin{align}\log_{\frac{1}{2}}(23)&=\frac{\ln(23)}{\ln(1/2)}\\&=\frac{3.135494}{-0.693147}\\&=-4.523562\end{align}$$
\(0.6309298 +2\log_3(x)\)
$$\begin{align} \log_3(2x^2)&=\log_3(2)+\log_3(x^2)& \text{multiplication rule}\\&=\frac{\ln(2)}{\ln(3)}+\log_3(x^2)&\text{change of basis used to rewrite} \log_3(2)\\&=0.6309298+2\log_3(x)& \text{exponent rule to simplify the second log}\end{align}$$
Note that we used the change of base formula simply to calculate a log of a number so we could get a decimal rather than being left with \(\log_3(2)\). The change of basis formula isn’t helpful for much else aside from that in these problems.
\(3.321928+2\log_2(x)-\log_2(z)\)
$$\begin{align} \log_2\left(\frac{30x^2}{3z}\right)&=\log_2(30x^2)-\log_2(3z)&\text{division rule}\\&=\log_2(30)+\log_2(x^2)-\log_2(3)-\log_2(z)&\text{multiplication rule}\\&=\frac{\ln(30)}{\ln(2)}+\log_2(x^2)-\frac{\ln{3}}{\ln{2}}-\log_2(z)&\text{change of base formula used on logs with numbers inside}\\&=4.9069+2\log_2(x)-1.58496-\log_2(z)&\text{computed the fractions and used exponent rule for \(log_2(x^2)\)}\\&=3.321928+2\log_2(x)-\log_2(z)&\text{combined the numbers}\end{align}$$
We’re only using the change of base formula to compute the logs with only numbers inside so we get a decimal result for those.
\(-2.31739+x\)
Note that there is one more step than the video shows: \(\log_5(3)=\frac{\ln(3)}{\ln(5)}=0.68261\). So the TRUE final answer is \(0.68261+x-3=-2.31739+x\)
$$1.16096+\frac{3}{2}\log_2(r)-\frac{5}{2}\log_2(z)$$
Then, using change of base formula we get
$$\begin{align}\frac{1}{2}\cdot\frac{\ln(5)}{\ln(2)}+\frac{3}{2}\log_2(r)-\frac{5}{2}\log_2(z)&=1.16096+\frac{3}{2}\log_2(r)-\frac{5}{2}\log_2(z)\end{align}$$
Note that the 1.16096 came from multiplying the \(1/2\) by the result of \(\frac{\ln{5}}{\ln{2}}\).