Exponential Equations (1 Viewer)

[qwerty44]

How do you solve algebraically besides trial and error?

 

[fortune_cookie]

That is the only way. Either trial and error, or by graphical methods (graph y=2^x and y=1+2x-x^2 accurately on graph paper and approximate the intersections).

In general there is no way to get an exact solution.

Of course in this case x=1. However, most of the time the solution will not be an integer.

 

[qwerty44]

That is the only way. Either trial and error, or by graphical methods (graph y=2^x and y=1+2x-x^2 accurately on graph paper and approximate the intersections).

There is no way to get an exact solution.

Of course, in this case x=1. However, most of the time the solution will not be an integer.

Thought so. When I came across these that is what I though then I saw a Cambridge question asking to show that they intersect at x=1 and 0. So I should just sub the x values they give to 'show'?

 

[fortune_cookie]

Thought so. When I came across these that is what I though then I saw a Cambridge question asking to show that they intersect at x=1 and 0. So I should just sub the x values they give to 'show'?

Whenever they say "show" you can just sub in the values to see that it is satisfied.

Same is true for simultaneous equations.

Say they gave you the lines 2x+3y=1 and y=3-4x and asked you to show that the intersection is (4/5 , -1/5)

You could just sub the values into both equations and show that they are both satisfied, which is much quicker than physically solving the equations, and it is a sufficent answer if the question says "show". .

 

[qwerty44]

Whenever they say "show" you can just sub in the values to see that it is satisfied.

Same is true for simultaneous equations.

Say they gave you the lines 2x+3y=1 and y=3-4x and asked you to show that the intersection is (4/5 , -1/5)

You could just sub the values into both equations and show that they are both satisfied, which is much quicker than physically solving the equations, and it is a sufficent answer if the question says "show". .

yes that makes sense because every equation with exponentials like OP has asked to show. thanks

 

[D94]

That is the only way. Either trial and error, or by graphical methods (graph y=2^x and y=1+2x-x^2 accurately on graph paper and approximate the intersections).

In general there is no way to get an exact solution.

Of course in this case x=1. However, most of the time the solution will not be an integer.

...for 2U Maths.

If this was MX1/MX2, then you can use Newton's Method.

 

[fortune_cookie]

...for 2U Maths.

If this was MX1/MX2, then you can use Newton's Method.

Once again, in general that would not give an exact solution.

 

[D94]

Once again, in general that would not give an exact solution.

True, I never said otherwise. I was disputing what you said about Trial and Error being the "only way". I believe, in an exam situation, Newton's Method would be the most accurate. (obviously, in this case, it is conveniently a calculable integer, so by observation would be the best method)

 

[fortune_cookie]

True, I never said otherwise. I was disputing what you said about Trial and Error being the "only way". I believe, in an exam situation, Newton's Method would be the most accurate. (obviously, in this case, it is conveniently a calculable integer, so by observation would be the best method)

Newton's Method is essentially a form of "trial and error" :|.

 

[iSplicer]

How do you solve algebraically besides trial and error?

You can't solve that algebraically, I'm pretty sure it's a transcendental equation (someone with uni maths experience may want to correct me on this).

Use newton's method/graphing/interval method if not blatant, thoughtless trial and error.

 

[Sy123]

I was able to algebraically simplify the expression into its simplest form of x=5- 2*2^x
This is done by doing ln to both sides after you make the RHS a perfect square, then playing around with log laws and the numbers after that.

 

[Sanjeet]

^ can you show that?

 

[Sy123]

The answer is much more evident there, and moreover, it is more evident that there is only 1 solution here since it is a linear function intersecting an exponential graph (in my simplified version)

 

[AAEldar]

The answer is much more evident there, and moreover, it is more evident that there is only 1 solution here since it is a linear function intersecting an exponential graph (in my simplified version)

How does





?

 

[Sy123]

Well, if you take it this way, ln of anything is always a value that is fixed, so whatever x is fixed at, ln is fixed. So in that line of working, Im basically saying that:

2 of the LHS = 1 of the RHS

Therefore 1 of the LHS = 0.5 of the RHS. Because its an equation and balance needs to be retained

I myself am a tiny bit unsure about that method, but my brain tells me its ok to do that. Or is that way just simply mathematically incorrect?

 

[iSplicer]

The answer is much more evident there, and moreover, it is more evident that there is only 1 solution here since it is a linear function intersecting an exponential graph (in my simplified version)

Intuitive idea, but I'm not sure if your line 6 is correct =)

 

[Sanjeet]

Well, if you take it this way, ln of anything is always a value that is fixed, so whatever x is fixed at, ln is fixed. So in that line of working, Im basically saying that:

2 of the LHS = 1 of the RHS

Therefore 1 of the LHS = 0.5 of the RHS. Because its an equation and balance needs to be retained

I myself am a tiny bit unsure about that method, but my brain tells me its ok to do that. Or is that way just simply mathematically incorrect?

you mean 1 of the LHS = 2 of the RHS
Your line 6 is wrong.

 

[fortune_cookie]

You also took logarithms of (x-1)^2

This is zero for the obvious solution x=1

ln(0) is not defined.

 

[Sy123]

Yes, I just graphed my solution and the question on Geogebra, and they intersect at y=0
i.e. I graphed y=5-2^(x+1)-x and y=1+2x-x^2-2^x

I thought it was fine though because I tested x=1, the solution

*embarassed*

 

[fortune_cookie]

lol and x=0 is also a solution. I cannot believe I missed that.