Once you understand The Relationship, you can use it to evaluate logarithmic expressions exactly; that is, you can use The Relationship to simplify the log expression to get one simple numerical value.

Note: For doing these evaluations, you will need to be sure that you're good with exponents, so you can recognize things like 216 being equal to 6^{3}.

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(You might not have known that fact but, trust me, you're gonna become very familiar with it, among many others.)

To evaluate (or simplify) a logarithmic expression without a calculator (when this simplification is possible), follow these steps:

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- Set the log expression equal to a placeholder variable. For instance, when asked to simplify log
_{3}(81), set this equal to, say,*x*, yielding the log equation log_{3}(81) =*x*. - Convert this log equation to the equivalent exponential equation. In our example, this would be the exponential equation 3
^{x}= 81. - Restate the numerical value in terms of the base. Continuing our example, this means finding the power that turns 3 into 81, which is 4.
- Equate the coefficients. In our case, this means taking 3
^{x}= 81 = 3^{4}and equating the powers, giving us*x*= 4.

(Note: The above only works when the log's argument is some integer or fractional power of the base. If the argument isn't a nice power of the base [or some power of a root of the base, about which more later], then you'll have moved into calculator-based evaluation. But while you're in the section where you're doing calculator-free simplifications, it is safe to assume that every argument will indeed be come well-behaved exact power of the base.)

If you don't know the particular power for a number that (you know from the log) has a certain base, then grab your calculator and start dividing by the base. Keep track of how many times you've divided, to find the number of factors of that base were used to get to that number.

In the case of the number 216 and a log with base 6, you'd divide 216 by 6 and get 36, which you know equals 6^{2}. This tells you that 216 = 6^{3}.

- Simplify log
_{2}(8)

This log is equal to some number, which I'll call *y*. This naming gives me the equation log_{2}(8) = *y*. Then The Relationship says:

2* ^{ y}* = 8

That is, log_{2}(8), also known as *y*, is the power that, when put on 2, will turn 2 into 8. The power that does this is 3:

2^{3} = 8

Since 2* ^{ y}* = 8 = 2

log_{2}(8) = 3

(Note: Setting the log expression equal to the variable *y*, which we found to equal the exponent on 6, illustrates how logs are exponents.

(For me, though, the "logs are powers" thing has always been more of an intellectual point than a meaningful help in understanding or working with logs. But if your instructor or textbook mentions this equivalence, the fact that *y* equalled the power on 2 is what they're talking about.)

- Simplify log
_{5}(25)

I'll start by setting the log equal to the variable *y*.

The Relationship says that, since log_{5}(25) = *y*, then:

5* ^{ y}* = 25

I know that 25 = 5^{2}. This means that:

5^{ y} = 5^{2}

*y* = 2

And *y* stands for the value of the log expression, so my hand-in answer is:

log_{5}(25) = 2

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While many of the expressions you'll be given to simplify will have whole-number values, exponents can be fractional, where fractional powers equate to roots. For instance, the power ½ on a base corresponds to the square root of that base. You will have exercises which require this knowledge.

- Simplify log
_{64}(4)

The Relationship says that this log represents the power *x* that, when put on 64, turns it into 4:

log_{64}(4) = *x*

64^{x} = 4

Remembering that 4^{3} is 64, remembering that fractional exponents correspond to roots, and noting that the cube root of 64 is 4, then:

64^{(1/3)} = 4

log_{64}(4) = *x* = 1/3

So my hand-in answer is:

log_{64}(4) = ^{1}/_{3}

This last example highlights the fact that, to be able to work intelligently with logs, you need to be pretty good with your exponents. So take the time to review them, if you're feeling a little shaky.

Sometimes, it will be difficult to figure out the exponents that you need. In many of these cases, it will be helpful to convert both the base and the argument into exponential expressions using a smaller base.

- Evaluate log
_{9}(27)

I start as usual, by naming the expression and converting the result to exponential form:

log_{9}(27) = *x*

9^{x} = 27

Hmm... The number 27 is not a power of 9. So what do I do?

I note that each of 27 and 9 is a power of 3, and I'll see where this leads:

9^{x} = 27

(3^{2})^{x} = 3^{3}

3^{2x} = 3^{3}

2*x* = 3

I'd assigned *x* to be the value of the original logarithm, so my answer is:

If you're not sure of this answer, then do the evaluation of the exponential. The one-half power means the square root, so you're taking the square root of 9, which is 3, and then cubing this result, which gives you 27.

- Simplify log
_{6}(6)

The Relationship says that, since log_{6}(6) = *y*, then 6* ^{ y}* = 6. The only power that doesn't change anything is the power 1. This means that:

6 = 6^{1}

6* ^{ y}* = 6

This means that *y* = 1, so my hand-in answer is:

log_{6}(6) = 1

This is always true: log_{b}(b) = 1 for *any* base b, not just for b = 6.

- Simplify log
_{3}(1)

The Relationship says that, since log_{3}(1) = *y*, then 3* ^{ y}* = 1. The only power that changes the base to 1 is zero.

This means that:

1 = 3^{0}

3* ^{ y}* = 3

*y* = 0

Then my hand-in answer is:

log_{3}(1) = 0

This is always true: log_{b}(1) = 0 for *any* base b, not just for b = 3.

You should expect to see "trick" questions like the folloowing on the next test:

- Simplify log
_{4}(−16)

The Relationship says that, since log_{4}(−16) = *y*, then 4* ^{ y}* = −16.

But wait! What power *y* could possibly turn a *positive* 4 into a *negative* 16? This just isn't possible, so my answer is:

no solution

This is always true: log_{b}(*a*) is undefined for *any* negative argument *a*, regardless of what the base is.

- Simplify log
_{2}(0)

The Relationship says that, since log_{2}(0) = *y*, then:

2* ^{ y}* = 0

But wait! What power *y* could possibly turn a 2 into a zero? This just isn't possible, so the answer is:

no solution

This is always true: log_{b}(0) is undefined for *any* base b, not just for b = 2.

- Simplify log
_{b}(b^{3})

The Relationship says that "log_{b}(b^{3}) = *y*" means "b^{ y} = b^{3}". Then clearly *y* = 3, so:

log_{b}(b^{3}) = 3

This is always true: log_{b}(b^{n}) = *n* for *any* base b.

Some students like to think of the above simplification as meaning that the b and the log-base-b "cancel out". This is not technically correct, but it can be a useful way of thinking of things.

(Just don't say it out loud in front of your instructor.)

- Simplify 2
^{log2(9)}

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Remember that a logarithm is just a power; granted, it is in this case a painfuly lumpy and long way of writing the power, but it is just a power, nonetheless.

The expression log_{2}(9) technically means "the power which, when put on 2, turns 2 into 9." And they've put that power onto 2, which means that the 2 has been turned into 9.

Looking at it another way, the expression 2^{log2(9)} = *y* means "log_{2}(*y*) = log_{2}(9)", which is the equivalent logarithmic statement, so *y* = 9. But *y* = 2^{log2(9)}, so 2^{log2(9)} = 9.

While the first way (using the "logs are just powers" definition) is technically correct, I find the second way (the "set the log expression equal to a variable, and convert the equation to exponential form" method) to be more intuitive and understandable. Either way, though, I get an answer of:

2^{log2(9)} = 9

This last example probably looks very complicated, and can feel quite confusing. But some students view the above problem as the 2 and the log-base-2 as "cancelling out". This is not technically correct, but can be a useful way of remembering how this type of problem works.

(But, again, don't say it out loud in front of your instructor.)

To synopsize, these are the take-aways from this lesson so far:

- The Relationship: "log
_{b}(*x*) =*y*" means the same thing as "b^{ y}=*x*". - Logarithms are really exponents (that is, orders or powers); they're just written differently.
- log
_{b}(b) = 1, for any base b, because b^{1}= b. - log
_{b}(1) = 0, for any base b, because b^{0}= 1. - log
_{b}(*a*) is undefined if*a*is negative. - log
_{b}(0) is undefined for any base b. - log
_{b}(b) =^{n}*n*, for any base b. - b
^{logb(a)}= a for any base b.

You might find it wise to do extra homework exercises, to make sure you have a good feel for the above facts.

URL: https://www.purplemath.com/modules/logs2.htm

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