y - ln x = 8

3x + y = 11

 

How do you solve it algebraically?

Try to solve for y with the second equation, plug it into the first. Move ln x to a side by itself and "e base" both sides of the equation. Solve for x. Plug x into the first equation solve for y.

What do you mean e base the whole thing?

I got to 3 - 3x = ln x

Do you mean e^(3-3x) = x

yes

But how do you solve it? It ends up being e^3 = x e^(3x)

hmmm...I;m sorry I'm stuck too. There has got to be a better way to do it. I graphed the functions, and got the answer x=1 and y=8 but I can't figure it out algebraically.

I simplified it to 1=(1/3)lnX+X, where X is less than or equal to 1, and greater than 0. That gives you a basic range and you could use logic. You know that if you use a decimal you'll have a negative with the natural log and when added to itself it can't add up to 1, so you could figure out that it has to be 1... I'm confused how to solve it algebraically though

I'm surprized none of the math people on this forum has helped out with this problem, yet. Can anyone help psi20 with this problem?

y - lnx = 8

3x + y = 1

 

y=(8+lnx)

 

3x+ (8+lnx) = 1

 

=> 3x+lnx=3....(1)

 

exponential of both sides of (1):

[math]e^{3x+lnx} = e^3[/math]

equate coefficients x=1

 

if x=1, y-ln1=8

so y-0=8

y=8

Do you just use logic to figure it out? Why do you know x=1 from 3x + lnx=3?

Seeing that x = 1 is a fairly trivial solution of the equation. Knowing it's the only solution is a bit of a beast though; the way I thought of doing it was to look at the function f(x) = 3x + log(x) - 3, then plot f(x) on a graph and see where it intersects the x-axis (it only does it at one place). It's not really a "proof" though.

It's not possible to solve this algebraically, if by that you mean using the usual 'elementary' functions. You can solve it using a computer package because it reduces to a non-elementary function called the LambertW function. This function is useful precisely because it allows one to solve these sorts of equations. It is a function that is one step beyond logs in the same way as logs are one step beyond polynomials.

 

If we start with [math]3x+lnx=3[/math] then [math]e^{3-3x}=x[/math] which can be rewritten as [math]3e^3=3xe^{3x}[/math] or [math]3e^3=we^w[/math] where [math]w=3x[/math]

 

w is the value of the LambertW function, which, as I've said isn't on a calculator, but is in most computer maths packages and you find [math]x=\frac{1}{3}\text{LambertW}(3e^3)[/math] which gives 1.

 

Don't understand or (hopefully) want to learn more? Then there's an excellent beginners guide to LambertW (including why it's called that) in the American Scientist at Why W?

Seeing that x = 1 is a fairly trivial solution of the equation. Knowing it's the only solution is a bit of a beast though; the way I thought of doing it was to look at the function f(x) = 3x + log(x) - 3, then plot f(x) on a graph and see where it intersects the x-axis (it only does it at one place). It's not really a "proof" though.

the easiest way to prove it would be to just differentiate f

 

f'(x) = 3 + 1/x

 

f' is positive wherever f is defined, so f is one-to-one

I was trying to show it in easier terms that everyone would understand. If I wanted to prove it formally I'd use the MVT and that fact above.

I was trying to show it in easier terms that everyone would understand. If I wanted to prove it formally I'd use the MVT and that fact above.

 

Can you state the mean value theorem please?

Let [math]f:[a,b] \to \mathbb{R}[/math]. If f is differentiable on (a,b) and continuous on [a,b] then there exists [math]c \in (a,b)[/math] such that:

 

[math]f'© = \frac{f(b) - f(a)}{b-a}[/math]

Let [math]f:[a' date=b] \to \mathbb{R}[/math]. If f is differentiable on (a,b) and continuous on [a,b] then there exists [math]c \in (a,b)[/math] such that:

 

[math]f'© = \frac{f(b) - f(a)}{b-a}[/math]

 

Thanks

 

Let me ask you this, how do you remember that?

 

Let f denote a function that maps the closed interval [a,b] onto the reals. If the function f is differentiable on the open interval (a,b), and continuous on the closed interval [a,b] then there is at least one point c, in the open interval (a,b) such that the derivative at c is equal to

 

f(b)-f(a)/(b-a)

I just have a nack for remembering theorems. I don't know how to explain it.

 

I just have a nack for remembering theorems. I don't know how to explain it.

 

 

I took advanced calc' date=' there was a picture for this one...

 

Don't worry about it...

 

Here I found it.

 

Mean Value Theorem

 

Ok that is easy to remember.

 

Tangent is parallel to secant, at at least one point c.