Solving Log Equations with Exponentials
The second type of log equation requires the use of The Relationship:
y = bx
...........is equivalent to............
(means the exact same thing as)
logb(y) = x
In animated form, the two equations are related as shown below:
Note that the base in both the exponential form of the equation and the logarithmic form of the equation is "b", but that the x and y switch sides when you switch between the two equations. If you can remember this — that whatever had been the argument of the log becomes the "equals" and whatever had been the "equals" becomes the exponent in the exponential, and vice versa — then you should not have too much trouble with solving log equations.
Since this equation is in the form "log(of something) equals a number", rather than "log(of something) equals log(of something else)", I can solve the equation by using The Relationship:
log2(x) = 4
24 = x
16 = x
I can solve this by converting the logarithmic statement into its equivalent exponential form, using The Relationship:
But 8 = 23, so I can equate powers of two:
Note that this could also have been solved by working directly from the definition of a logarithm.
What power, when put on "2", would give you an 8? The power 3, of course!
If you wanted to give yourself a lot of work, you could also do this one in your calculator, using the change-of-base formula:
Plug this into your calculator, and you'll get "3" as your answer. While this change-of-base technique is not particularly useful in this case, you can see that it does work. (Try it on your calculator, if you haven't already, so you're sure you know which keys to punch, and in which order.) You will need this technique in later problems.
I'm not saying that you'll necessarily want to solve equations using the change-of-base formula, or always by using the definition of logs, or any other particular method. But I am suggesting that you should make sure that you're comfortable with the various methods, and that you shouldn't panic if you and a friend used totally different methods for solving the same equation.
Solve log2(x) + log2(x – 2) = 3
I can't do anything with this equation yet, because I don't yet have it in the "log(of something) equals a number" form. So I'll need to use log rules to combine the two terms on the left-hand side of the equation:
log2(x) + log2(x – 2) = 3
log2[(x)(x – 2)] = 3
log2(x2 – 2x) = 3
Now the equation is arranged in a useful way. At this point, I can use The Relationship to convert the log form of the equation to the corresponding exponential form, and then I can solve the result:
log2(x2 – 2x) = 3
23 = x2 – 2x
8 = x2 – 2x
0 = x2 – 2x – 8
0 = (x – 4)(x + 2)
x = 4, –2
But if x = –2, then "log2(x)", from the original logarithmic equation, will have a negative number for its argument (as will the term "log2(x – 2)"). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be x = –2.
Then my solution is:
Keep in mind that you can always check your answers to any "solving" exercise by plugging those answers back into the original equation and checking that the solution "works". In this case, I'll plug my solution value into either side of the original equation, and verify that each side evaluates to the same number:
the left-hand side:
log2(x) + log2(x – 2)
= log2(4) + log2(4 – 2)3
= log2(4) + log2(2)
= log2(22) + log2(21)
= 2 + 1 = 3
The right-hand side of the original equation was already simplified to "3", so this solution checks.
This equation may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice. I start with the original equation and work with the "outer" log:
The Relationship converts the above to:
Now I'll apply The Relationship a second time:
Then the solution is:
Solve log2(x2) = (log2(x))2
First, I'll expand the square on the right-hand side to be the explicit product of two logs:
log2(x2) = [log2(x)]2
log2(x2) = [log2(x)] [log2(x)]
Then I'll apply the log rule to move the "squared" from inside the log on the left-hand side of the equation, taking it out in front of that log as a multiplier:
2·log2(x) = [log2(x)] [log2(x)]
Then I'll move that term from the left-hand side of the equation to the right-hand side:
0 = [log2(x)] [log2(x)] – 2·log2(x)
This equation may look bad, but take a close look. It's nothing more than a factoring exercise at this point. So I'll factor, and then I'll solve the factors by using The Relationship:
0 = [log2(x)] [log2(x) – 2]
log2(x) = 0 or log2(x) – 2 = 0
20 = x or log2(x) = 2
1 = x or 22 = x
1 = x or 4 = x
Then my solution is:
You can use the Mathway widget below to practice solving logarithmic equations (or skip the widget and continue with the lesson). Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.
(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)
Log Calculator (Logarithm)
Please provide any two values to calculate the third in the logarithm equation logbx=y. It can accept "e" as a base input.
What is Log?
The logarithm, or log, is the inverse of the mathematical operation of exponentiation. This means that the log of a number is the number that a fixed base has to be raised to in order to yield the number. Conventionally, log implies that base 10 is being used, though the base can technically be anything. When the base is e, ln is usually written, rather than loge. log2, the binary logarithm, is another base that is typically used with logarithms. If, for example:
x = by; then y = logbx; where b is the base
Each of the mentioned bases is typically used in different applications. Base 10 is commonly used in science and engineering, base e in math and physics, and base 2 in computer science.
Basic Log Rules
When the argument of a logarithm is the product of two numerals, the logarithm can be re-written as the addition of the logarithm of each of the numerals.
logb(x × y) = logbx + logby
EX: log(1 × 10) = log(1) + log(10) = 0 + 1 = 1
When the argument of a logarithm is a fraction, the logarithm can be re-written as the subtraction of the logarithm of the numerator minus the logarithm of the denominator.
logb(x / y) = logbx - logby
EX: log(10 / 2) = log(10) - log(2) = 1 - 0.301 = 0.699
If there is an exponent in the argument of a logarithm, the exponent can be pulled out of the logarithm and multiplied.
logbxy = y × logbx
EX: log(26) = 6 × log(2) = 1.806
It is also possible to change the base of the logarithm using the following rule.
|EX: log10(x) =|
To switch the base and argument, use the following rule.
|EX: log5(2) =|
Other common logarithms to take note of include:
logb(1) = 0
logb(b) = 1
logb(0) = undefined
limx→0+logb(x) = - ∞
ln(ex) = x
Log Base 2 Calculator
As mentioned at the end of the above section, the binary logarithm is a special case of the logarithmic function with base. That means that we'll have expressions of the form , and we'll ask ourselves to what power we should raise in order to obtain . For instance, we can easily observe that .
Seemingly, is a number like any other. However, it has some interesting properties. E.g., it is the smallest prime number and the only even one. Moreover, it's the base for any computer-related operations via the binary representation.
Since it's so important, let's recall a few basic powers of. Remember that the exponent can also be or even negative.
Now we can see some more examples than just the from above. For instance, we can say that the log with base of is . Similarly, or .
But what is, say,? Surely, is not a power of .
To be precise, it's not an integer power of. We have to remember that there are also fractional exponents, and indeed, here, we need one of those. Unfortunately, they're not so simple to guess. In some cases, we can try to use tricks like the change of base formula, but, in general, it's best to use an external tool - something like our log base 2 calculator.
In it, you can see two variable windows: and . Hopefully, the notation is self-explanatory. For example, if you'd like to find , you need to input under , and the calculator will give you the answer in the other window. If you require , you enter . Also, note how Omni's log base 2 calculator works both ways: you can either input the value of and obtain or the other way round.
That'll be enough for today's lesson. Go, my young padawan, and make sure to play around with the calculator or any other algebra-related tool that we have on offer.
Logarithms: Simplifying with
"The Relationship" (page 2 of 3)
Sections: Introduction to logs, Simplifying log expressions, Common and natural logs
This log is equal to some number, which I'll call y. This naming gives me the equation log2(8) = y. Then the Relationship says:
That is, log2(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:
Since 2 y = 8 = 23, then it must be true that y = 3, and I get:
The Relationship says that, since log5(25) = y, then 5 y = 25. This means that the given log log5(25) is equal to the power y that, when put on 5, turns 5 into 25. The required power is 2, because 52 = 25. Then 52 = 5 y = 25, so:
- Simplify log64(4). Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
The Relationship says that this log represents the power y that, when put on 64, turns it into 4. Remembering that 43 is 64, and remembering that fractional exponents correspond to roots, this means that the cube root of 64 is 4, so 64(1/3) = 4. Then:
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.
The Relationship says that, since log6(6) = y, then 6 y = 6. But 6 = 61, so 6 y = 61, and y = 1. That is:
This is always true: logb(b) = 1 for any base b, not just for b = 6.
The Relationship says that, since log3(1) = y, then 3 y = 1. But 1 = 30, so 3 y = 30, and y = 0. That is:
This is always true: logb(1) = 0 for any base b, not just for b = 3.
The Relationship says that, since log4(16) = y, then 4 y = 16. But wait! What power y could possibly turn a positive4 into a negative16? This just isn't possible, so the answer is:
This is always true: logb(a) is undefined for any negative argumenta, regardless of what the base is.
The Relationship says that, since log2(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:
This is always true: logb(0) is undefined for any base b, not just for b = 2.
The Relationship says that "logb(b3) = y" means "b y = b3". Then clearly y = 3, so:
This is always true: logb(bn) = nfor 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.
Remember that a logarithm is just a power; it's a lumpy and long way of writing the power, but it's just a power, nonetheless. The expression "log2(9)" 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, "2log2(9) = y" means "log2(y) = log2(9)" (which is the equivalent logarithmic statement), so y = 9. But y = 2log2(9), so 2log2(9) = 9.
While the second way is technically correct, I find the first way to be more intuitive and understandable. Either way, though, I get an answer of:
This last example probably looks very complicated, and, in the technical explanation, it is. Look instead at the intuitive explanation (in the first paragraph). Some students even view the above problem as the 2 and the log-base-2 as "cancelling out", which is not technically correct, but can be a useful way of remembering how this type of problem works.
To synopsize, these are the things you should know from this lesson so far:
- The Relationship: "logb(x) = y" means the same thing as "b y = x".
- Logarithms are really exponents (powers); they're just written differently.
- logb(b) = 1, for any base b, because b1 = b.
- logb(1) = 0, for any base b, because b0 = 1.
- logb(a) is undefined if a is negative.
- logb(0) is undefined for any base b.
- logb(bn) = n, for any base b.
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Cite this article as:
Stapel, Elizabeth. "Logarithms: Simplifying with 'The Relationship'." Purplemath. Available from
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