Today's soundtrack is Petra: Petra Praise: The Rock Cries Out, the first Petra Praise album. Released in 1989, it's an arena-rock take on what were at that time contemporary worship songs; it was revolutionary for its day!
This afternoon, I'm learning about the Laws of Logarithms.
Power Rule
The first law is the Law of Logarithms for Powers, which applies to logarithms for any base. It states that logₐxⁿ = nlogₐx. The restrictions are as follows: x∈ℝ, n∈ℝ, x > 0, a > 0, and a ≠ 1.
We can use the Power Rule to solve exponential equations. This is especially handy in cases where it is difficult to convert the two sides' arguments to the same base. Here's how to do it:
How to use the Power Rule to solve exponential equations
Take the log of both sides
3ˣ = 14 becomes log3ˣ = log14
Apply the Power Rule to bring down the exponents
log3ˣ = log14 becomes xlog3 = log14
Isolate the variable
Divide both sides by the log value on the side where the variable is
(xlog3 = log14) / log3 cancels log3 out on the left, giving us x = (log14/log3)
Solve using technology
log14/log3 = 2.402173503; thus, x = 2.402173503
We can basically the same method to quickly find values of uncommon logarithms. If given log19=x, we could simply type in [LOG](19) in our calculator and be done with it. But what if we are asked to find log₄19=x? We can't just type a four into the equation; it doesn't work. What we do instead is divide the logarithm of the argument by the logarithm of the base (log19/log4). This concept leads us to the next rule, the Change of Base Rule for Logarithms, which states that logₓa=(log₁₀a/log₁₀x). This rule allows us to find the logarithm of an uncommon base. Let's see how it works:
How to use the Change of Base Rule to find the logarithm of an uncommon base
Convert from logarithmic form to exponential form
log₄19=x becomes 19=4ˣ
Take the log of both sides
log4ˣ = log19
Bring down the exponent(s)
xlog4 = log19
Isolate the variable through division
[(xlog4 = log19) / log4] = [x=(log19/log4)]
Next, let's look at the Law of Logarithms for Multiplication. This law tells us that logₐxy=logₐx+logₐy. So, if we are asked to find the solution to log(3.2 x 4), it will be the same solution as we would get for log3.2 + log4. Once again, our restrictions apply: a > 0, and a ≠ 1.
If asked to write log8 as a sum of two logarithms, we know that two logarithms added together equal a single multiplied logarithm. 8 can be divided by 4 and 2; thus, log4 + log2 = log8.
In the same way, if asked to turn log4 + log2 into a single logarithm, our answer would be log8.
Now we'll move on to the Law of Logarithms for Division. Very similar to the law above, it tells us that logₐx/y=logₐx-logₐy, with a > 0, and a ≠ 1.
If asked to write log8 as a difference of two logarithms, we know that a logarithm subtracted from another logarithm equals a single rational logarithm. The quotient of 16/2 is 8; thus, log16 - log2 = log8.
In the same way, if we are asked to convert log16-log2 into a single logarithm, our answer is log8.
The Reciprocal Rule says that logₐx=1/logₓa. We can prove this by applying the Change of Base rule and then reversing it again.
Our last rule, the Extended Power Rule, tells us that logₐnbⁿ = logₐb. This means that if our base and our argument have the same exponent, the exponents cancel out. But if our base is a reciprocal and we need to flip it, we will apply a negative to the exponent of a.
We will now finish up this post by looking at the logarithm properties.
Logarithm Properties
logₐaˣ=x
We've seen this in action many times: log10 = 1, log₇7 = 1. With both of those examples, the exponent was 1. So log10⁴ = 4, and log log₇7⁴ = 4 also.
aˡᵒᵍₐˣ=x
If we see logarithms in the exponent position, we can quickly simplify them.
If the two as are not the same value, the answer will be xⁿ, with n representing the value that the small a must be raised to exponentially to match the large a. For example, in the equation 8^log₂x, 2³ = 8, so our answer is x³.
If a number precedes log, then we must apply that exponent to our answer. So 5^2log₅x = x².
That's it for today; next time, we'll be learning how to solve exponential and logarithmic equations!