- Intro to logarithm properties (1 of 2)
- Intro to logarithm properties (2 of 2)
- Intro to logarithm properties
- Using the logarithmic product rule
- Using the logarithmic power rule
- Use the properties of logarithms
- Using the properties of logarithms: multiple steps
- Proof of the logarithm product rule
- Proof of the logarithm quotient and power rules
- Justifying the logarithm properties
Sal rewrites log₅(x³) as 3log₅(x). Created by Sal Khan and Monterey Institute for Technology and Education.
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- How do you do log2 -1(25 votes)
- I'd like to see a step by step solution to the following equation: 5^(3x+2)=3^(4x-1) solve for x(10 votes)
- log 5^(3x+2) = log3^(4x-1)
(3x+2)log5 = (4x-1)log3
3xlog5 + 2log5 = 4xlog3 - log3
3xlog5 - 4xlog3 = -2log5 - log3
x(3log5 - 4log3) = -(2log5+log3)
x = -log75 / (3log5 -4log3)(41 votes)
- what's log2(2x) - log2x^3 = 5 ?(7 votes)
- simple just follow these steps!
1)log2(2x)-log2(x^3)=5 write in the form that is logbX-logbY=logbX/Y;
then this will become log2(2x/ (x^3) )=5
3) 2^5=2/x^2 => 32=2/x^2
= x=1/4 .....final answer;
- What happens if you have logarithms involving imaginary or complex numbers?(7 votes)
- Since you can't use normal methods to evaluate the log, you would need to resort to things like Euler's Formula and Demovire's Theorem.(10 votes)
- at about3:30, the instructor states that (a^b)^d=a^bd. How is it possible for the exponent to shift down a degree?(4 votes)
- This is how exponents work. It doesn't shift it down a degree. Here are a few cases and examples:
This is not unlike what happens when you multiply two numbers with the same base raise to different powers. In that case you add the exponents:
Good question. It is confusing at first, but when you get the hang of it, it's not too hard.(6 votes)
- how would you find x in 36^x=6?(3 votes)
- Start by making both sides have a common base.
36 can be written as 6^2 and remember 6=6^1. This changes the equation into:
(6^2)^x = 6^1
Now the only way for these 2 sides to be equal is for the exponents to be equal. So: 2x must equal 1
Solve 2x =1
x = 1/2
Hope this helps.(6 votes)
- does this rule apply to when we have the whole logarithm raised to a power, instead of just the argument? Ex:log_3(x^3)=3log_3(x), (log_3(x))^3=3log_3(x).(3 votes)
- In general, (log_b(a))^p will not equal p * log_b(a). The power rule with logarithms only applies when the logarithm input is raised to a power.(4 votes)
- Having a problem what is log(1000)= 4.605? This problem doesn’t seem to make sense to me and many of the problems following are similar(3 votes)
- log (1000) = 4.605 is an equation, not an expression. You can't solve an equation without any variables. Did you meant that the base of the logarithm was unknown?(3 votes)
- how to do log2^x+log3^81=1(2 votes)
- Does this mean: log base 2 of x plus log base 3 of 81 equals 1?
I will write that like this: log_2(x)+log_3(81)=1
We should start by finding log_3(81): log_2(x)+4=1
Next, subtract 4 from both sides: log_2(x)=-3
log_2(x)=-3 is the same as 2^-3=x.
Keep going!(2 votes)
- I came up with another prove for this rule:
C^A=C*C*C...(the product of A number of C)
according to the logarithmic product rule we learnt from last lesson,
logB(C)^A=logb(C)+logb(C)+... (A number of logb(C))
so it equals to A*logbC(3 votes)
We're asked to simplify log base 5 of x to the third. And once again, we're just going to rewrite this in a different way. You could argue whether it's going to be more simple or not. And the logarithm property that I'm guessing that we should use for this example right here is the property-- if I take log base x of-- let me pick some more letters here, log base x of y to the zth power. This is the same thing as z times log base x of y. So this is a logarithm property. If I'm taking the logarithm of a given base of something to a power, I could take that power out front and multiply that times the log of the base, of just the y in this case. So we apply this property over here. And in a second, once I do this problem, we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties. But if we just apply that over here, we get log base 5 of x to the third. Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as-- let me do this in a different color-- that 3 is the same thing-- we could put it out front-- that's the same thing as 3 times the logarithm base 5 of x. And we're done. This is just another way of writing it using this property. And so you could argue that this is a what-- maybe this is a simplification because you took the exponent outside of the logarithm, and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that-- and I'll just pick some arbitrary letters here-- let's say that we know that a to the b power is equal to c. And so if we know that-- that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b. To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here, and raise it to the dth power. So let me make it-- so let's raise-- take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting this statement. But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a B. Actually, let's say I'm using all lowercase letters. This is a lowercase c. So let me write it this way, a to the-- so I'm going to raise this to the dth power, and I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now, what's interesting over here is we can now say-- what we could do is we can use what we know about exponent properties. Say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. This is-- let me do that in a different color. I've already used that green. This right over here, using what we know about exponent properties, this is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we would write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to get to c to the dth power? To get to this? I have to raise it to the bd power. But what do we know that b is? We already know that b is this thing right over here. So if we substitute this in for b, and we can rewrite this as db, we get logarithm base a of c to the dth power is equal to bd, or you could also call that db, if you switch the order. And so that's equal to d times b. b is just log base a of c. So there you have it. We just derived this property. Log base a of c to the dth, that's the same thing as d times log base a of c, which we applied right over here.