You have learned various rules for manipulating and simplifying expressions with exponents, such as the rule that says that *x*^{3} × *x*^{5} equals *x*^{8} because you can add the exponents.

There are similar rules for logarithms.

(I'll provide proofs for each of the rules. You almost certainly don't need to know these proofs, so feel free to ignore those parts.)

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The Three Basic Log Rules:

1) Product Rule: log_{b}(*mn*) = log_{b}(*m*) + log_{b}(*n*)

2) Quotient Rule: log_{b}(^{m}/_{n}) = log_{b}(*m*) – log_{b}(*n*)

3) Power Rule: log_{b}(*m** ^{n}*) =

In less formal terms, these three log rules might be expressed as:

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1) Product Rule: Multiplication inside the log can be turned into addition outside the log, and vice versa.

2) Quotient Rule: Division inside the log can be turned into subtraction outside the log, and vice versa.

3) Power Rule: An exponent on everything inside a log can be moved out front as a multiplier, and vice versa.

Warning: Just as when you're dealing with exponents, the above rules work *only* if the bases are the same. For instance, the expression "log_{d}(*m*) + log_{b}(*n*)" cannot be simplified, because the bases (the **d** and the **b**) are not the same, just as *x*^{2} × *y*^{3} cannot be simplified because the bases (the *x* and *y*) are not the same.

Below are some examples of these log rules at work, using the base-10 common log:

1) log(2*x*) = log(2) + log(*x*)

log(2) + log(3) = log(2 × 3) = log(6)

2) log(*x*/3) = log(*x*) − log(3)

log(12) − log(3) = log(12 ÷ 3) = log(4)

3) log(*x*^{2}) = 2 × log(*x*)

3 × log(5) = log(5^{3}) = log(125)

It's all good and well to state these relationships. But how do we know that these rule are true? (If you don't care — which is fine — continue to the rest of the rules.)

I'll start with proving the Power Rule first, because I'll be using the Power Rule to prove the Product and Quotient Rules. (In the following proofs, *x* and *y* are placeholders with unspecified values, that I'm using to make my life easier.)

3) Power Rule: Let log_{b}(a^{n}) = *x* (for some unknown value *x*) and let log_{b}(a) = *y* (for some unknown value *y*). By the definition of logarithms (being the reverse of exponentials), we get b^{x} = a^{n} and b^{y} = a. Then:

b^{x} = a^{n} = (b^{y})^{n} = b^{yn}

This means that b^{x} = b^{yn}, so clearly *x* = *y*n. We can convert the *x* back to log_{b}(a^{n}), so:

log_{b}(a^{n}) = *y*n

And we can replace the *y* on the right-hand side of the above equation with log_{b}(a), which gives us:

log_{b}(a^{n}) = [log_{b}(a)](n) = n × log_{b}(a)

This proves the Power Rule.

With the Power Rule proven, I can now prove the Product and Quotient Rules.

1) Product Rule: Let log_{b}(m) = *x* and log_{b}(n) = *y*, so b^{x} = m and b^{y} = n. Then:

(b^{x})(b^{y}) = b^{x+y} = mn

Taking the log of the middle and right-hand terms of the above equation, we get:

log_{b}(b^{x+y}) = log_{b}(mn)

The Power Rule turns the left-hand side into (*x* + *y*)log_{b}(b). Letting log_{b}(b) equal *z*, we get b^{z} = b = b^{1}, so *z* = 1 and log_{b}(b) = 1. Now we can back-substitute for *x* and *y*, replacing them with the expressions for which they were named:

*x* + *y* = log_{b}(m) + log_{b}(n)

Putting it all together, we have:

log_{b}(mn) = log_{b}(b^{x+y})

= *x*+*y* = log_{b}(m) + log_{b}(n)

This proves the Product Rule. The same sort of reasoning will prove the Quotient Rule.

2) Quotient Rule: Let log_{b}(m) = *x* and log_{b}(n) = *y*. Then b^{x} = m and b^{y} = n.

Take the log of the first and last portions of the above equality (that is, omitting the middle portion) to get:

Now back-substitute for *x* and *y* (and again omit the middle portion) to get:

This proves the Quotient Rule.

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In your algebra class, you'll use the log rules to "expand" and "condense" logarithmic expressions. The expanding is what I did in the first in each pair of examples above; the condensing is the second in each pair.

However, in other classes (such as calculus), you will need these rules to simplify complicated log expressions andconvert painful log expressions into nicer forms that will allow other mathematics to be done.

There are additional log rules:

Additional Log Rules:

4) Power-of-1 Rule: log_{b}(b) = 1

5) Power-of-0 Rule: log_{b}(1) = 0

6) Log-Cancelling Rule; log_{b}(b^{n}) = n

7) Power-Cancelling Rule: b^{logb(n)} = n

8) Change-of-Base Formula:

9) Base-Switch Rule:

Note that, in all cases, the logarithm's base b must be positive and not equal to 1, and all values inside logarithms must be positive. (If you want to go straight to using these rules, skipping their proofs, then continue on to the next page.)

Rules (4) and (5) above are based on the fact that anything raised to the power 1 is just itself, and anything raised to the power 0 is just 1:

4) Power-of-1 Rule: b^{1} = b means the same thing as log_{b}(b) = 1

5) Power-of-0 Rule: b^{0} = 1 means the same thing as log_{b}(1) = 0

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Rules (6) and (7) are based on the definition of logs as being themselves exponents. (You don't actually "cancel" anything, but it kinda feels like you do, hence my names for these rules.)

6) Log-Cancelling Rule: Let log_{b}(b^{n}) = *x* (for some unknown value *x*). This means the same thing as b^{x} = b^{n}. Then obviously *x* = n, so log_{b}(b^{n}) = n.

7) Power-Cancelling Rule: Let b^{logb(n)} = *x* (for some unknown value *x*). This means the same thing as log_{b}(*x*) = log_{b}(n). Then obviously *x* = n, so b^{logb(n)} = n.

The Change-of-Base Formula can (and will) be incredibly helpful to you when you're working with logs on your calculator. You may be asked to evaluate a log expression where the log's base is something other than 10 or *e*. But your calculator can evaluate *only* logs with one of these two bases. The Change-of-Base Formula gives you a way out: you can convert the log expression into an equivalent expression that uses one of the two bases that your calculator is able to handle.

To prove the Change-of-Base Formula, I'll use the same techniques of renaming as I used in the proofs above. The steps will be a bit more involved, though.

8) Change-of-Base Formula: Let log_{b}(a) = *z*, log_{c}(a) = *y*, and log_{c}(b) = *x*. Then b^{x} = a, c^{y} = b, and c^{x} = a.

c^{y} = b ⇒ b^{z} = c^{yz}

c^{x} = a = b^{z} = c^{yz}

Since c^{x} = c^{yz}, then *x* = *yz*. Dividing through, we get . Back-substituting, we then get:

This proves the Change-of-Base Formula.

The only remaining rule is the Base-Switch Rule.

9) Base-Switch Rule: Let log_{b}(a) = *x*. Then b^{x} = a. Taking the base-a log of both sides, we get:

log_{a}(b^{x}) = log_{a}(a)

*x* log_{a}(b) = 1

Note: Before I'd checked online for all the rules for logs (to make sure I wasn't forgetting anything), I'd never even heard of the Base-Switch Rule. I don't know where you'd use it.

URL: https://www.purplemath.com/modules/logrules.htm

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