Sal solves the equation log(x)+log(3)=2log(4)-log(2). Created by Sal Khan and Monterey Institute for Technology and Education.
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- Where are logs used in real life?(9 votes)
- 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.
- 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.(15 votes)
- Order of operations still apply, but remember the properties, in this case
Go learn yourself the properties, they are very useful :)(15 votes)
- I'm getting confused at3:42, 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.(9 votes)
- 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(5 votes)
- How do you solve a logarithmic equation with the same base on both sides?
Example: loga(b+1) = loga(c+1)
- 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)(2 votes)
- When you have constants on each side of the equation and they are being raised to binomial powers, how do you solve?
For example: 5^(x+3) = 7^(2x+3)(2 votes)
- Problems such as these have infinitely many solutions if you allow nonreal complex values of x. Your class most likely won't cover the nonreal complex solutions as that is rather advanced math. So, assuming only real values of x, here is how you would solve a problem like this:
5^(x+3) = 7^(2x+3)
ln [ 5^(x+3)] = ln [7^(2x+3) ]
(x+3) ln (5) = (2x+3) ln (7)
Now use the distributive property:
x ln( 5) + 3 ln 5 = 2x ln(7) + 3 ln(7)
Rearrange to get x on the left hand side:
x ln 5 - 2x ln 7 = 3 ln 7 - 3 ln 5
Factoring gives us:
x (ln 5 - 2 ln 7) = 3 (ln 7 - ln 5)
x = 3 (ln 7 - ln 5) / (ln 5 - 2 ln 7)
There are many ways of expressing this solution that might not look too much like this. For example, if I had used 5 as the base of my log instead of the natural log I would get this as the final answer:
3 [ log₅ (7) - 1 ]/[ 1 − 2 log₅(7) ]
The two answers are equal, even though they don't look much alike.(4 votes)
- so if i had, 3log(base2)x - log(2)(x-2) = 4
ive really got log2(x^3/x-2) = 4?(2 votes)
- I'm slightly confused at3:16of 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 :)(3 votes)
- 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.(1 vote)
- Does anybody know how to solve the equation 2^(x+4)=64x. I keep getting stuck.(2 votes)
- Did you, by any chance, miss-type that. Are you sure it's not supposed to be 2^(x+4)=64? With no x?(1 vote)
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.