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## Algebra 2 (Eureka Math/EngageNY)

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

Lesson 7: Topic B: Lessons 13-15: Logarithms and equations- Solving exponential equations using logarithms: base-10
- Solving exponential equations using logarithms
- Solve exponential equations using logarithms: base-10 and base-e
- Solving exponential equations using logarithms: base-2
- Solve exponential equations using logarithms: base-2 and other bases
- Logarithmic equations: variable in the argument
- Logarithmic equations: variable in the base

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# Logarithmic equations: variable in the argument

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?

- 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.

http://en.wikipedia.org/wiki/Music_and_mathematics(42 votes)

- 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

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 :)(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)

Thanks!(2 votes)- ... Don't you think
`b`

is equal to`c`

at the first sight?(9 votes)

- 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)
- Yes you would. Same for radical equations.(5 votes)

- What is y=b to the 2nd power(3 votes)
- 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?(2 votes)- 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.(3 votes)

- so if i had, 3log(base2)x - log(2)(x-2) = 4

ive really got log2(x^3/x-2) = 4?(2 votes)- That looks right, yes. Further, you have 2^4, or 16, equal to x^3/(x-2).(3 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)

## 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.