What is a logarithm?
Exponentials and logarithms are related by the general log rule
When we first start learning about logarithms, it’s helpful to think about how they’re related to exponents, since exponents are something we already understand.
Exponents vs. logarithms
Remember that an exponent tells us how many times to multiply the base by itself. In other words, in ???2^3???, the exponent of ???3??? tells us to multiply the base ???2??? by itself three times. And that tells us how to solve for the value of ???2^3???:
???2^3=2\cdot2\cdot2=8???
We already know that this is what exponents do. What we haven’t learned yet is what to do when we have this instead:
???2^x=8???
The logarithm is what we would use to find the value of ???x??? in this expression, because the logarithm lets us solve for the value of an exponent. Logarithms tell us how many times we multiply one number by itself in order to get a different number. So when we already have the number that we’re multiplying by itself, ???2???, and we have the result, ???8???, logarithms (or “logs,” for short) tell us what the value of the exponent needs to be in order to make the equation true.
To change ???2^x=8??? into a log, we write
???\log_2{(8)}=3???
In both ???2^3=8??? and ???\log_2{(8)}=3???, the ???2??? is the base, the ???8??? is the argument, and the ???3??? is the exponent. So we’ll read ???\log_2{(8)}=3??? as
“log base ???2??? of ???8??? is ???3???,” or
“the log of ???8??? with base ???2??? is ???3???,” or
“the base-???2??? log of ???8??? is ???3???”
Realize that the “base” of a log is the same as the “base” of an exponent. In other words, in both ???2^3=8??? and ???\log_2{(8)}=3???, the ???2??? is the base. Remembering that can help us when we’re converting back and forth between logs and exponents.
Not multiplication
It’s also worth pointing out that, in the equation ???\log_2{(8)}=3???, ???\log_2??? is not multiplied by ???(8)???. It’s tempting to think that the parentheses indicate multiplication, but they don’t. Instead, the big number that comes after the base is always the “argument” of the log function.
Think back to function notation, where we talked about functions like ???f(x)=2x+1???. Remember that the function notation ???f(x)??? doesn’t indicate ???f??? multiplied by ???x???. Instead, it means that “the function ???f??? is being evaluated at ???x???,” or that “???f??? is a function of ???x???.”
And the same is always true for logarithms. In ???\log_2{(8)}=3???, the log function is being evaluated at ???8???.
The general log rule
This basic log rule that relates exponents to logs we can write as
Given the equation ???a^x=y???,
the associated log is ???\log_a{(y)}=x???,
and vice versa.
Let’s do an example where we convert from one form to the other.
How to use the general log rule to convert between exponentials and logarithms
Take the course
Want to learn more about Algebra 2? I have a step-by-step course for that. :)
Converting an exponential equation into a logarithmic equation
Example
Convert ???5^4=625??? to its logarithmic form.
If we match up ???5^4=625??? to ???a^x=y??? from the general log rule, we can say
???a=5???
???x=4???
???y=625???
Plugging these values into ???\log_a{(y)}=x??? from the general log rule, we get
???\log_a{(y)}=x???
???\log_5{(625)}=4???
Realize that we could have just as easily started with ???\log_5{(625)}=4???, identified ???a???, ???x???, and ???y???, and converted to form ???a^x=y???.
In both forms, the expression means “multiplying ???5??? four times gives ???625???.”
Let’s do an example where we solve a log equation for an unknown variable.
Example
Solve the equation.
???\log_2{(16)}=x???
This log equation is asking “How many times do you have to multiply ???2??? in order to get ???16????” Using the general log rule, we can rewrite the log as
???2^x=16???
We know that ???2^4??? gives
???2^4=2\cdot2\cdot2\cdot2=16???
so the solution to the log equation is ???x=4???.
We can solve log problems regardless of which value is unknown. In the last example, ???\log_2{(16)}=x???, the result was the unknown value. If instead the base were the unknown value, we would’ve had
???\log_x{(16)}=4???
???x^4=16???
???x=2???
Or if instead the argument were the unknown value, we would’ve had
???\log_2{(x)}=4???
???2^4=x???
???x=16???