moved - Intersection between linear and exponential equations

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Thepotatom

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My problem is the following: Let the functions: f(x)= 2^(x+3) -5 and g(x)=3x+3
a)For what values of x is f(x)=g(x)?

I tried to solve for x, but I really cannot, is it possible to solve without Desmos?

Thanks,
Thomas Charland
 
Thepotatom said:
My problem is the following: Let the functions: f(x)= 2^(x+3) -5 and g(x)=3x+3
a) For what values of x is f(x)=g(x)?

I tried to solve for x, but I really cannot, is it possible to solve without Desmos?

Thanks,
Thomas Charland
Click to expand...

1) Without Desmos? Yes. There are many ways to draw graphs.

2) There are many Numerical Methods that do not involve graphing.
 
Thepotatom said:
My problem is the following: Let the functions: f(x)= 2^(x+3) -5 and g(x)=3x+3
a)For what values of x is f(x)=g(x)?

I tried to solve for x, but I really cannot, is it possible to solve without Desmos?

Thanks,
Thomas Charland
Click to expand...
It is great that you tried to solve this problem. Can you please show us at least one of your attempts so we can offer you some guidance?
 
Sure. the furthest I've got is to reduce it to: 2^x= 3x/8 +1
But then I don't really know to to find x since there is one on both sides of the equation.
What should I do next?
 
Thepotatom said:
My problem is the following: Let the functions: f(x)= 2^(x+3) -5 and g(x)=3x+3
a)For what values of x is f(x)=g(x)?

I tried to solve for x, but I really cannot, is it possible to solve without Desmos?

Thanks,
Thomas Charland
Click to expand...

The equation they want you to solve, f(x) = g(x), is 2^(x+3) -5 = 3x+3. This is a transcendental equation, which can't be solved by algebraic methods; but as was mentioned, there are "numerical methods" to find an approximate solution to any equation (and sometimes they accidentally find an exact solution).

The simplest numerical method is "guess and check". Just try some numbers and see if they make it true -- for example, try x = 0, 1, 2, ... . Then try simple negative numbers. If one of those is a solution, you're ready for the next step. If not, you can try to narrow down a solution, by finding numbers between which there has to be a solution, and then trying numbers that are closer together. (The more advanced numerical methods just make that process more efficient by giving you a good guess for the next step each time.)

Of course, that can't tell you whether you have found all the solutions. One way to do that is to sketch the graph yourself (e.g. using calculus to find maxima and minima), and use what you discover in that process to decide whether there can be other solutions, and where they would be. You don't need to be dependent on technology to do your graphing for you! (Incidentally, just graphing by plotting points can amount to guess-and-check, so you might want to start there.)

EDIT: I've given you a better hint here than you may realize. Be sure to try what I've suggested!
 
Last edited:
Thepotatom said:
Sure. the furthest I've got is to reduce it to: 2^x= 3x/8 +1
But then I don't really know to to find x since there is one on both sides of the equation.
What should I do next?
Click to expand...
What method/s have been taught to you?

What was the topic being covered in class when this was given as HW?

Have you been taught to use Newton's method to approximate roots of non-linear equations?
 
Subhotosh Khan, We are seeing the exponential function in class and we usually only solve equation like: 3^x= 3/5 with (log3/5)/log3 We have not learned more advanced methods for approximating like the previous reply was suggesting.
 
Thepotatom said:
Subhotosh Khan, We are seeing the exponential function in class and we usually only solve equation like: 3^x= 3/5 with (log3/5)/log3 We have not learned more advanced methods for approximating like the previous reply was suggesting.
Click to expand...

That's right. Variables IN exponents. Get both IN and OUT and it's another story. It is a common disappointment of sometimes even very successful early algebra students. Something like, "What do you mean I CAN'T solve it?!" That word "can't" is the problem. Many want it to mean we're just not being sufficiently clever. This is one of the great moments in your mathematics career. Showing that things CAN be done is only part of the story. Sometimes, we manage to show that things CANNOT be done. It is a beautiful thing.
 
Thepotatom said:
Sure. the furthest I've got is to reduce it to: 2^x= 3x/8 +1
But then I don't really know to to find x since there is one on both sides of the equation.
What should I do next?
Click to expand...
Instead of always jumping into using some formula I ALWAYS try to see if I can guess the answer, especially for a transcendental equation like this one. You really should see a solution to 2^x= 3x/8 +1. Non-zero raised to 0 is 1 and a multiple of 0 (like (3/8)*0) is 0. Does that help you? Now there may be other solutions. Try graphing to see if that looks like a possibility.

Here is an example where I always think first and truly benefit by it. Find the equation of the line passing through (4,8) and (12,24). I notice that the y value is always twice the x value. So y = 2x. This is so much quicker and more fun doing then finding the slope and then the y intercept. Always think for obvious answers for at least a few seconds. It really pays off in the end.
 
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