So does order of operation still have to be followed with logarithms? Here's why I ask this: 2*log(4) - log(2) What I did was I first used the division property and I got 2*log(4/2) = 2*log(2). Only after this I moved the 2 in front to be the exponent of log(2) so I got log(4). In the video Sal first multiplied and then divided the logarithm, resulting in log(8). Have I done something wrong? Thanks.

Order of operations still apply, but remember the properties, in this case a*log_b(c)=log_b(c)^a 2*log(4)-log(2) log(4)^2-log(2) log(16)-log(2) log(16/2) log(8) Go learn yourself the properties, they are very useful :)

Where are logs used in real life?

The human ear works as a logarithmic function. The tempered musical scale is exponential so after passing through a logarithmic function (ear) it become linear. This mix of functions makes the transition from notes of the scale perceived by our brain softly as if the notes were located exactly one after the other. Basically, the frequencies of the musical notes are equally logarithmic scaled. http://en.wikipedia.org/wiki/Music_and_mathematics

I'm getting confused at 3:42, when he offers an alternate way of solving the equation. Can someone theoretically explain to me how raising both sides of the equation to the log's base solves the problem? I understand the lesson up until that part.

logarithms are just inverse functions of exponential functions so that the base and the exponents cancel and equal 1 .try this logany base (withthat number)=1 as well exponets leading coeffitient with raised with any logsame numbe =1 let say 10^x(power)=100 by logarithm rules it inverse it intern of x log(10_base)(100)=x so that x=2 log( 10^x(power))=log(100) this simplifies to x=log 100 or 2

How do you solve a logarithmic equation with the same base on both sides? Example: loga(b+1) = loga(c+1) Thanks!

... Don't you think `b` is equal to `c` at the first sight?

When solving for a logarithm, would you need to find the domain of the equation as well? (To ensure that the solution is within the domain of the equation)

Yes you would. Same for radical equations.

What if instead of having log_ throughout the equation, you have two logarithms on one side but just a number on the other side of the equal sign? Ex] log base of 9 (x -7) plus log base of 9 (x - 7) is equal to one. Would you multiply the (x - 7) with each other and then equal it to one?

Here's how to solve that kind of problem. Let me know if you don't understand any step and I'll try to explain in greater detail. log₉(x-7) + log₉(x-7) = 1 2log₉(x-7) = 1 log₉(x-7) = ½ 9^[log₉(x-7)] = 9^(½) x-7 = 3 x = 10 Here is a similar but slightly more difficult problem: log₉(x-4) + log₉(x+4) = 1 log₉{(x-4)(x+4)} = 1 9^[log₉{(x-4)(x+4)} ] = 9^1 (x-4)(x+4) = 9 x²-16 = 9 x² = 25 x = 5 Note: we have to discard any values for x that would lead to a log with an argument that is 0 or a negative number, that is why I only used the principle square roots. In some occasions, you may need to use the negative square roots -- it all depends on what keeps the argument a positive number.

so if i had, 3log(base2)x - log(2)(x-2) = 4 ive really got log2(x^3/x-2) = 4?

That looks right, yes. Further, you have 2^4, or 16, equal to x^3/(x-2).

Main content

Algebra 2 (Eureka Math/EngageNY)

Course: Algebra 2 (Eureka Math/EngageNY) > Unit 3

Lesson 7: Topic B: Lessons 13-15: Logarithms and equations

Logarithmic equations: variable in the argument

Name: Logarithmic equations: variable in the argument
Uploaded: 2011-11-10T11:00:27Z
Description: Sal solves the equation log(x)+log(3)=2log(4)-log(2).

Google Classroom

Sal solves the equation log(x)+log(3)=2log(4)-log(2). Created by Sal Khan and Monterey Institute for Technology and Education.

Want to join the conversation?

Sort by:

Jason Swearingen
Posted 11 years ago. Direct link to Jason Swearingen's post “Where are logs used in re...”
Where are logs used in real life?
Button navigates to signup pageComment on Jason Swearingen's post “Where are logs used in re...”
(9 votes)
Answer
- Marcos Vinicius
  Posted 11 years ago. Direct link to Marcos Vinicius's post “The human ear works as a ...”
  The human ear works as a logarithmic function. The tempered musical scale is exponential so after passing through a logarithmic function (ear) it become linear. This mix of functions makes the transition from notes of the scale perceived by our brain softly as if the notes were located exactly one after the other. Basically, the frequencies of the musical notes are equally logarithmic scaled.
  http://en.wikipedia.org/wiki/Music_and_mathematics
  Button navigates to signup page
  (42 votes)
Emanuel
Posted 11 years ago. Direct link to Emanuel's post “So does order of operatio...”
So does order of operation still have to be followed with logarithms? Here's why I ask this:

2*log(4) - log(2)

What I did was I first used the division property and I got 2*log(4/2) = 2*log(2). Only after this I moved the 2 in front to be the exponent of log(2) so I got log(4).

In the video Sal first multiplied and then divided the logarithm, resulting in log(8).

Have I done something wrong? Thanks.
Button navigates to signup pageComment on Emanuel's post “So does order of operatio...”
(15 votes)
Answer
- Kristoffer Naterstad Seljestad
  Posted 11 years ago. Direct link to Kristoffer Naterstad Seljestad's post “Order of operations still...”
  Order of operations still apply, but remember the properties, in this case
  
  a*log_b(c)=log_b(c)^a
  
  2*log(4)-log(2)
  log(4)^2-log(2)
  log(16)-log(2)
  log(16/2)
  log(8)
  
  Go learn yourself the properties, they are very useful :)
  Button navigates to signup page
  (15 votes)
MiguelDominguez520
Posted 11 years ago. Direct link to MiguelDominguez520's post “I'm getting confused at 3...”
I'm getting confused at
3:42
, when he offers an alternate way of solving the equation. Can someone theoretically explain to me how raising both sides of the equation to the log's base solves the problem?
I understand the lesson up until that part.
Button navigates to signup pageButton navigates to signup page
(9 votes)
Answer
- metuxxman
  Posted 11 years ago. Direct link to metuxxman's post “logarithms are just inver...”
  logarithms are just inverse functions of exponential functions so that the base and the exponents cancel and equal 1 .try this logany base (withthat number)=1
  as well exponets leading coeffitient with raised with any logsame numbe =1
  
  let say 10^x(power)=100 by logarithm rules it inverse it intern of x
  log(10_base)(100)=x so that x=2
  log( 10^x(power))=log(100) this simplifies to x=log 100 or 2
  Button navigates to signup page
  (5 votes)
Arshiya Jain
Posted 9 years ago. Direct link to Arshiya Jain's post “How do you solve a logari...”
How do you solve a logarithmic equation with the same base on both sides?
Example: loga(b+1) = loga(c+1)
Thanks!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Cyan Wind
  Posted 9 years ago. Direct link to Cyan Wind's post “... Don't you think `b` i...”
  ... Don't you think b is equal to c at the first sight?
  Button navigates to signup page
  (9 votes)
Catherine
Posted 7 years ago. Direct link to Catherine's post “When solving for a logari...”
When solving for a logarithm, would you need to find the domain of the equation as well? (To ensure that the solution is within the domain of the equation)
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- andrewp18
  Posted 7 years ago. Direct link to andrewp18's post “Yes you would. Same for r...”
  Yes you would. Same for radical equations.
  Comment on andrewp18's post “Yes you would. Same for r...”
  (5 votes)
Carmen Corley
Posted 5 years ago. Direct link to Carmen Corley's post “What is y=b to the 2nd po...”
What is y=b to the 2nd power
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
malinali1997
Posted 9 years ago. Direct link to malinali1997's post “What if instead of having...”
What if instead of having log_ throughout the equation, you have two logarithms on one side but just a number on the other side of the equal sign?
Ex]
log base of 9 (x -7) plus log base of 9 (x - 7) is equal to one.
Would you multiply the (x - 7) with each other and then equal it to one?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Just Keith
  Posted 9 years ago. Direct link to Just Keith's post “Here's how to solve that ...”
  Here's how to solve that kind of problem. Let me know if you don't understand any step and I'll try to explain in greater detail.
  log₉(x-7) + log₉(x-7) = 1
  2log₉(x-7) = 1
  log₉(x-7) = ½
  9^[log₉(x-7)] = 9^(½)
  x-7 = 3
  x = 10
  Here is a similar but slightly more difficult problem:
  log₉(x-4) + log₉(x+4) = 1
  log₉{(x-4)(x+4)} = 1
  9^[log₉{(x-4)(x+4)} ] = 9^1
  (x-4)(x+4) = 9
  x²-16 = 9
  x² = 25
  x = 5
  Note: we have to discard any values for x that would lead to a log with an argument that is 0 or a negative number, that is why I only used the principle square roots. In some occasions, you may need to use the negative square roots -- it all depends on what keeps the argument a positive number.
  Button navigates to signup page
  (3 votes)
Benjamin Duyvesteyn
Posted 8 years ago. Direct link to Benjamin Duyvesteyn's post “so if i had, 3log(base2)x...”
so if i had, 3log(base2)x - log(2)(x-2) = 4
ive really got log2(x^3/x-2) = 4?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- George Winslow
  Posted 8 years ago. Direct link to George Winslow's post “That looks right, yes. Fu...”
  That looks right, yes. Further, you have 2^4, or 16, equal to x^3/(x-2).
  Button navigates to signup page
  (3 votes)
Laura Londono
Posted 9 years ago. Direct link to Laura Londono's post “I'm slightly confused at ...”
I'm slightly confused at
3:16
of the video. When i got to the stage of: log3x = log 8
i divided log3 on both sides leaving:

x = log8 / log3

I didn't drop the log on both sides as he did in the video, and when i put it in the calculator, they're two different answers. Is dividing out by log3 agebraicly incorrect or is it just assumed you have to drop the log when they're on both sides of the equation in the above situation?

Thanks so much for the video though! It was super helpful :)
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Rashel
  Posted 9 years ago. Direct link to Rashel's post “Yes, when you have the sa...”
  Yes, when you have the same log base on both sides of the equation (log base 10 or log base 2) they will cancel each other out. It's like if you had x+2=x+2. The x's will cancel out when you subtract the x's from each other. When solving logs, many of the same rules for solving equations apply, the only difference is in addition to solving for x, you have to condense the logs.
  Button navigates to signup page
  (1 vote)
Matt Samaro
Posted 10 years ago. Direct link to Matt Samaro's post “Does anybody know how to ...”
Does anybody know how to solve the equation 2^(x+4)=64x. I keep getting stuck.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Derek Muller
  Posted 10 years ago. Direct link to Derek Muller's post “Did you, by any chance, m...”
  Did you, by any chance, miss-type that. Are you sure it's not supposed to be 2^(x+4)=64? With no x?
  Button navigates to signup page
  (1 vote)

Video transcript

We're asked to solve the log of x plus log of 3 is equal to 2 log of 4 minus log of 2. So let me just rewrite it. So we have the log of x plus the log of 3 is equal to 2 times the log of 4 minus the log of 2, or the logarithm of 2. And this is a reminder. Whenever you see a logarithm written without a base, the implicit base is 10. So we could write 10 here, 10 here, 10 here, and 10 here. But for the rest of this example, I'll just skip writing the 10 just because it'll save a little bit of time. But remember, this literally means log base 10. So this expression, right over here, is the power I have to raise 10 to to get x, the power I have to raise 10 to to get 3. Now with that out of the way, let's see what logarithm properties we can use. So we know, if we-- and these are all the same base-- we know that if we have log base a of b plus log base a of c, then this is the same thing as log base a of bc. And we also know-- so let me write all the logarithm properties that we know over here. We also know that if we have a logarithm-- let me write it this way, actually-- if I have b times the log base a of c, this is equal to log base a of c to the bth power. And we also know, and this is derived really straight from both of these, is that if I have log base a of b minus log base a of c, that this is equal to the log base a of b over c. And this is really straight derived from these two right over here. Now with that out of the way, let's see what we can apply. So right over here, we have all the logs are the same base. And we have logarithm of x plus logarithm of 3. So by this property right over here, the sum of logarithms with the same base, this is going to be equal to log base 3-- sorry, log base 10-- so I'll just write it here. log base 10 of 3 times x, of 3x. Then, based on this property right over here, this thing could be rewritten-- so this is going to be equal to-- this thing can be written as log base 10 of 4 to the second power, which is really just 16. So this is just going to be 16. And then we still have minus logarithm base 10 of 2. And now, using this last property, we know we have one logarithm minus another logarithm. This is going to be equal to log base 10 of 16 over 2, 16 divided by 2, which is the same thing as 8. So the right-hand side simplifies to log base 10 of 8. The left-hand side is log base 10 of 3x. So if 10 to some power is going to be equal to 3x. And 10 to the same power is going to be equal to 8. So 3x must be equal to 8. 3x is equal to 8, and then we can divide both sides by 3. Divide both sides by 3, you get x is equal to 8 over 3. One way, this little step here, I said, look, 10 to the-- this is an exponent. If I raise 10 to this exponent, I get 3x, 10 to this exponent, I get 8. So 8 and 3x must be the same thing. One other way you could have thought about this is, let's take 10 to this power, on both sides. So you could say 10 to this power, and then 10 to this power over here. If I raise 10 to the power that I need to raise 10 to to get to 3x, well, I'm just going to get 3x. If I raise 10 to the power that I need to raise 10 to to get 8, I'm just going to get 8. So once again, you've got the 3x is equal to 8, and then you can simplify. You get x is equal to 8/3.