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## Algebra 2

### Course: Algebra 2 > Unit 8

Lesson 3: Properties of logarithms- 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

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# Using the logarithmic power rule

Sal rewrites log₅(x³) as 3log₅(x). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- How do you do log2 -1(25 votes)
- This is undefined, because you cannot raise power to 2 in order to get -1.(6 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

2)log2(2/x^2)=5

3) 2^5=2/x^2 => 32=2/x^2

= 32x^2=2

= x^2=1/16

= x=1/4 .....final answer;

10Q(6 votes)

- 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:

(2^3)^2=8^2=64

(2^3)^2=2^(3*2)=2^6=64 (2^1=2,2^2=2,2^3=8,2^4=16,2^5=32,2^6=64)

(3^2)^3=9^3=729

(3^2)^3=3^(2*3)=3^6=729 (3,9,27,81,243,729)

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:

2^2*2^4=2^(2+4)=2^6=64

2^2*2^4=4*16=64

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

Simplify: 6^(2x)=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.

2^-3=1/8

x=1/8

Keep going!(2 votes)

- I came up with another prove for this rule:

Prove logb(C)^A=Alogb(C):

solution:

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)

## Video transcript

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.