Main content

## Algebra (all content)

### Course: Algebra (all content) > Unit 5

Lesson 8: Solving equations by graphing (Algebra 2 level)- Interpreting equations graphically
- Interpreting equations graphically (example 2)
- Interpret equations graphically
- Solving equations graphically (1 of 2)
- Solving equations graphically (2 of 2)
- Solving equations graphically
- Solve equations graphically

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Solving equations graphically (1 of 2)

CCSS.Math:

Sal solves the equation e^x=1/[x(x-1)(x-2)] by considering the graphs of y=e^x and y=1/[x(x-1)(x-2)]. Created by Sal Khan.

## Want to join the conversation?

- I don't get it... how is 7.846 and 7.633 within 0.01 ?(14 votes)
- Those are the values of E(x) and R(x). You want x within 0.01 of the actual value of x, not E(x) within 0.01 of R(x).(24 votes)

- At2:05Sal uses his calculator and chooses "e" and raises that to a power of 2.1. What is "e" and where did that irrational come from?(13 votes)
- e is a constant, just like PI is a constant. e is something like: 2.71828.... e is the base of natural logarithm. I believe that e stands for Euler's number.(18 votes)

- I have never learned this stuff. are there videos before this that will help me do this?(12 votes)
- This video gives an introduction to euler's number (e).

https://www.khanacademy.org/math/algebra2/exponential_and_logarithmic_func/continuous_compounding/v/compound-interest-and-e-part-2

This might help too (from wikipedia): The number e is an important mathematical constant that is the base of the natural logarithm. It is approximately equal to 2.71828, and is the limit of (1 + 1/n)n as n approaches infinity, an expression that arises in the study of compound interest. It can also be calculated as the sum of the infinite series.

https://en.wikipedia.org/wiki/E_(mathematical_constant)(2 votes)

- At the point Sal was looking for, the y values change enormously for tiny changes in x. Sal guesses a starting x of 2.1 by looking at the x axis. So would there be a way to use an estimate on the graph for the y axis, in order to go back and get a better starting point for the x value?(4 votes)
- If you used an inverse function for either R(x) or E(x) , then you could plugin an estimate of maybe 7.8 into that function, to come up with a starting value of x. e^x conveniently has an inverse function, which is ln(x).

ln(7.8) comes up as 2.05412373369554605284 on my calculator, which gives you a**very**good starting point, yes. :)(8 votes)

- What is the value of e at4:07?(3 votes)
- e is an irrational number like π (meaning it cannot be written as the ratio of 2 integers and thus in its decimal form it will go on forever without any pattern). The first few digits of e are:

e ≈ 2.718(3 votes)

- What is "e"? I've run into it, and I know it's aprox. 2.71812, but what is it's specific "job" in mathematics?(2 votes)
- The number represented by the variable 'e' is called Euler's Number. In math, it represents the base for something called the natural logarithm, which is a process that counteracts something being raised to a power, much like addition counteracts subtraction.(4 votes)

- Please, how exactly is E(x)=R(x) within 0.01 ?(3 votes)
- When it's much closer than either 2.05 and 2.07.(1 vote)

- I seems that the correct Algebra way to solve this would be to subtract e^x from 1/(x(x-1)(x-2) and set it equal to zero. i.e. 1/(x*(x-1)*x-2))-e^x=0. This is actually kind of what I did in Excel. I just made two rows of formula and subtracted the results. If I knew how to solve for x with that natural log (e) that is exactly what I would do.(2 votes)
- the solution suggested by jeffery works well

log natural in excel is ln()

charting in excel is as good as a calculator(2 votes)

- What does "within 0.01" mean exactly?(2 votes)
- That probably means the answer which you obtain in the end should be limited to the hundredths digit or 0.01...and thats what he did, he got the answer till the hundredths digit and did not solve it any further even though it is possible.(3 votes)

- Isn't this technically trial and error/Guess and Check?(2 votes)
- To some extent, yes. When you use graphing to find the solution to a system of equations, your answer will only be as good as the accuracy of your graphing. If you do the graph by hand, it is very likely that lines won't be straight, your scale might be off a little, etc. All these types of small deviations open up the potential for a close, but inaccurate solution. Thus, it is always a good ideal to check whatever solution you find to verify that it really is a solution.(1 vote)

## Video transcript

Graphs of e of x
equals e to the x and r of x is equal to 1 over x
times x minus 1 times x minus 2 are shown below. Estimate the solution to e
to the x-- so that's e of x-- being equal to essentially
r of x within 0.01. So we want to
figure out for what value does e of x equal r of x? And they want us to estimate it. We can either just try
to get as close as we can from this graph. They want us to be within 0.01. And we can also
use a calculator, kind of try numbers out
to hopefully zero in on this point right
over here, where e of x is equal to r of x. So what I want to do is, let
me draw a little table here. Let's try out some x values. And then for each
of these x values, let's see where we land on e of
x and where we land on r of x, and then we can decide whether
we are too high or too low. And I encourage you to pause
this video before I actually go ahead and do this, and
try to do this on your own. But I do suggest using
some form of calculator or, well, probably a calculator. I'm assuming you've
given a go at it, and now I will attempt it. Now just eyeballing
it-- and eyeballing it is helpful, because that'll
give us kind of our first order approximation of at what x value
are these two functions equal. If I just look at this
graph the way it's drawn, it looks like this is
pretty close to 2.1. It looks like when x is
2.1, both of those functions look pretty close
to-- I don't know. This looks like about 7.7 or
7.8 or something like that. But let's figure out
what they're doing. So let's see, when
x is equal to 2.1-- get my calculator out-- when
x is equal to 2.1, well, e of x is just e to the x power. So e to the 2.1 power
is equal to 8.166. Let me write that down, 8.166. And what is r of x? r of x is 1 divided by x,
so that's going to be 2.1. Times x minus 1. Well, that's going to be
1.1, so that's times 1.1. Times x minus 2. Well, that's just going
to be 0.1, times 0.1. And that is equal to
4-- did I do that right? No, that can't be. 2.1 over-- 2.1
times 1.1 times 0.1. 1 over all of that. 4.32? Let's see, 2.1 r
of x is 4 point-- I guess that's possible. Actually, that looks
right, because r of x declines so sharply
right over here. So it's actually, at 2.1
where actually r of x is actually closer to right
over here, give or take. So it's equal to 4.32. So 2.1 e of x is actually a
much larger value than r of x. So e of x is clearly
too high. r of x is already dropped
a good bit by then. If I were to go all
the way down to 2, at 2 it looks like actually r
of x kind of spikes up. It just goes to infinity
as we approach 2. So we're not going to go
all the way down to 2, but why don't we lower
this a little bit. Why don't we try 2.05? So 2.05, what is e of x? e of x is e to the x, right? So e to the 2.05 power gets
us 7.76-- I'll round it, 8-- 7.768. Approximately 7--
actually, all these are approximate, so
I'll just write 7.768. And what is r of x? I'll just keep rounding
to the thousands. Here, well, we didn't have
to round too much just because that was so far
off, but I'll put it there. Actually, it was
329, so I could-- let me write it this way-- 3290. So let me throw
that 9 here, just so everything-- we evaluate
the function to thousandths. So let's evaluate r of
x, when we're at 2.05, it's going to be 1 divided by
x, which is now 2.05, times x minus 1, which is 1.05, times
x minus 2, which is 0.05. And that gets us to 9.29-- I'll
round to 2-- 9.292, so 9.292. So now we're on this
side, where r of x is roughly right over
here and it's more than e of x, which is at 7.7,
which is right around here. So now our x value is too low. So maybe let's see if we
can go a little bit higher. And let's try to go roughly
halfway between these two, but I don't want to get too
precise, because you have to get to the nearest hundredth. So let's go to 2.07. So e to the 2.07 is equal to
7.925 if I round it, 7.925. I want to do all this in green
just to be color consistent. And now let's evaluate r
of x at that same value. So 1 divided by x, which is
2.07, times that minus 1, which is 1.07, times that
minus 2, which is 0.07, which gives us 6.44, I
guess we could say 6.450. So at 2.05 that was too
low, 2.07 is too high. Now, r of x has
dropped below e of x. So we know the right answer is
in between these two numbers, and so if we select
2.06 that's definitely going to be within 0.01
of the right answer. So I would go with
2.06 is definitely going to be within the 0.01 of
the correct solution to this. But just for fun, let's
actually just try it out. So e to the 2.06 is 7.84--
I guess we could round to 6. And if we were to
evaluate r of x, it's 1 divided by 2.06
times that minus 1, which is 1.06, times 0.06. It gets us to 7.632. So we're also
getting pretty close, but our precision
that they gave, they don't say that they have
to be within each other of that, they say, let's
estimate the solution. So there's some actual
precise solution to this right over here, some x
value, where these are actually equal to each other. That's the x value, which gives
us this point of intersection. We just have to get within
0.01 of that x value, and 2.06 definitely
does the trick.