Solvable Algebra equation?

[Gamer30]

Good day everyone,
I wish to seek all maths experts for help.
The question is:
3^x+2^x=5

I'm sure most of all know the value of X is 1 but what I'm interested is how to prove that the value of X is 1??
May I know how to show the working steps towards obtaining the value.

As such, I would like to kindly hope maths enthusiastic expertise can help me with this.
Do sincerely looking forward in helping me to solve the above equation .
Thanks in advance.

 

[The Highlander]

Good day everyone,
I wish to seek all maths experts for help.
The question is:
3^x+2^x=5

I'm sure most of all know the value of X is 1 but what I'm interested is how to prove that the value of X is 1??
May I know how to show the working steps towards obtaining the value.

As such, I would like to kindly hope maths enthusiastic expertise can help me with this.
Do sincerely looking forward in helping me to solve the above equation .
Thanks in advance.

The expression you have given above is: 3x² + 2^x = 5 but I suspect that what you meant to write was: 3x² + 2x = 5.

If I am correct then this equation (3x² + 2x = 5) is what is called a quadratic equation and there is more than one solution to it. There are in fact two possible solutions (only one of which is 1).

There are different ways to find the solution(s) of quadratic equations. The first step is usually to put it into the form: ax² + bx + c = 0 and then, if possible, to factorize that (or "complete the square").
If factorization is not possible then one may resort to use of the quadratic formula (be sure to read the whole webpage).

To begin, you would start by stating that:-

[math]3x^2+2x=5\implies 3x^2+2x-5=0[/math]
and you would then factorize that to find the two values of x that give zero as the answer.

Factorizing the equation should result in a new equation like this:-

[math](3x\qquad)(x\qquad)=0[/math]
Can you fill in the blank spaces inside the brackets with + or - signs and numbers such that when you multiply out the brackets you get back to the original equation ([imath]3x^2+2x-5=0[/imath])?
If not then I suggest you spend a bit more time reading through the linked webpages I have given you.

If, on the other hand, you really did mean to write 3x² + 2^x = 5 as the equation then, again, there is more than one possible value for x (again, only one of which is 1) but I don't propose to deal with that any further because I sincerely believe that you did mean to write 3x² + 2x = 5.

If I am wrong about that, then please let us know and further advice may be offered in that respect.

However, if I am correct in my assumption about the equation that you really want advice on then please read through the links I have provided and then attempt to factorize it using the hint I provided for you up above ([imath](3x\qquad)(x\qquad)=0[/imath]).

Hope that helps.

 

[blamocur]

I'm sure most of all know the value of X is 1 but what I'm interested is how to prove that the value of X is 1??

[imath]f(x) = 3^x+2^x[/imath] is a monotonic function, which means that for any [imath]a[/imath] the equation[imath]f(x) = a[/imath] can have only one solution.

 

[BigBeachBanana]

[imath]f(x) = 3^x+2^x[/imath] is a monotonic function, which means that for any [imath]a[/imath] the equation[imath]f(x) = a[/imath] can have only one solution.

@Gamer30 You might ask how to show that the function is monotonic. The answer lies in its derivative.

 

[Dr.Peterson]

Good day everyone,
I wish to seek all maths experts for help.
The question is:
3^x+2^x=5

I'm sure most of all know the value of X is 1 but what I'm interested is how to prove that the value of X is 1??
May I know how to show the working steps towards obtaining the value.

First, I thnk it's clear that you don't mean a quadratic equation, but exactly what you wrote, [imath]3^x+2^x=5[/imath], which is an exponential equation. You probably know how to solve a quadratic equation.

Second, you're asking two different questions, perhaps without realizing it.

If you're expecting algebraic manipulations to solve it (that is, to find the entire solution set), as in solving a linear or quadratic equation, that won't work. (Many students don't realize that most equations you can write can't be solved by the methods we teach in school!)

For this sort of equation, the best you can do to find the solution is either to graph the equation and see it, or to just recognize that 3+2=5 and know it in a flash of insight (which mathematicians call "solving by inspection"). This part, you've already done.

Then you can prove that it is a solution by putting x = 1 into the equation and showing that it is true. This, too, you have already done.

But that only finds a solution; you still need to prove that it is the complete solution. That is where monotonicity (that is, the fact that the function is always increasing, and therefore is one-to-one, comes in.

Now you need to tell us how much you know of algebra and calculus; the latter is the easiest (and maybe only) way to prove this.

 

[The Highlander]

First, I thnk it's clear that you don't mean a quadratic equation, but exactly what you wrote, [imath]3^x+2^x=5[/imath], which is an exponential equation. You probably know how to solve a quadratic equation.

Yes, I completely misread the OP!
My apologies for wasting space!

 

[Gamer30]

First, I thnk it's clear that you don't mean a quadratic equation, but exactly what you wrote, [imath]3^x+2^x=5[/imath], which is an exponential equation. You probably know how to solve a quadratic equation.

Second, you're asking two different questions, perhaps without realizing it.

If you're expecting algebraic manipulations to solve it (that is, to find the entire solution set), as in solving a linear or quadratic equation, that won't work. (Many students don't realize that most equations you can write can't be solved by the methods we teach in school!)

For this sort of equation, the best you can do to find the solution is either to graph the equation and see it, or to just recognize that 3+2=5 and know it in a flash of insight (which mathematicians call "solving by inspection"). This part, you've already done.

Then you can prove that it is a solution by putting x = 1 into the equation and showing that it is true. This, too, you have already done.

But that only finds a solution; you still need to prove that it is the complete solution. That is where monotonicity (that is, the fact that the function is always increasing, and therefore is one-to-one, comes in.

Now you need to tell us how much you know of algebra and calculus; the latter is the easiest (and maybe only) way to prove this.

Hi Dr. Peterson,
First, I would like to clarify that the question posted above is an exponential equation (not linear or quadratic equation).
If the methods we are taught in school can't be used to solve such equations, then may I know what kind of other ways would u propose in order to solve it? (I know "by inspection" is a quick way provided the power of the exponential equations is not complicated such as 3^x^x or other more complex ones)
Btw, may I know if graphing the equation also part of the working steps? (Awarded marks, I mean)
I'm not sure if my current knowledge of algebra and calculus is enough to solve such equations bcos I'm only a "high school" student (not university graduates). Therefore, I'm posting such equation so as to learn more in prep. for the future exams.
Hence, may I know what is the most easiest and "convincing" way to solve equations like these? Would greatly appreciate if u could list out all the methods that u know as I'm inclined to learn more from u.
My concern is to learn how to obtain a "complete" set of solution for questions like the above.
Once again, I hope u all could contribute and offer solutions for questions like these as this will benefit learners, not only students.
Thank you.

 

[Dr.Peterson]

Btw, may I know if graphing the equation also part of the working steps? (Awarded marks, I mean)

On an exam, you won't be given an equation that can't be solved by methods you have been taught; if you need some method like using a graphing tool, they would tell you it's allowed.

On the other hand, it's quite possible that the equation you asked about could be on an exam, because it tests your creativity in solving a problem without a standard method. As another example, you can solve 4^x - 3*2^x = 4 algebraically.

I'm not sure if my current knowledge of algebra and calculus is enough to solve such equations bcos I'm only a "high school" student (not university graduates). Therefore, I'm posting such equation so as to learn more in prep. for the future exams.
Hence, may I know what is the most easiest and "convincing" way to solve equations like these? Would greatly appreciate if u could list out all the methods that u know as I'm inclined to learn more from u.

The only major thing I didn't already mention (besides specific algebraic methods like using logs to solve simpler exponential equations) is what are called numerical methods; these (such as Newton's method) are used when such equations arise in real life (where we can't prevent them!). Possibly some are taught in high schools today; again, you could see whether that is mentioned in descriptions of what is covered by a particular exam. And the question would have to mention something about approximation, which would be a cue that such methods are expected.

(Fixed - thanks, @blamocur)

 

[blamocur]

you can solve 4^x - 3*2^2 = 4 algebraically.

4^x - 3*2^x = 4 ?

 

[Gamer30]

4^x - 3*2^x = 4 ?

Hi, can u show how to solve it algebraically? Would greatly appreciate it.
Thanks.

 

[fresh_42]

Hi, can u show how to solve it algebraically? Would greatly appreciate it.
Thanks.

I like to say: Remove what disturbs you most! This is clearly [imath] 2^x [/imath] in that case, so we substitute [imath] y=2^x. [/imath] Can you solve the equation [imath] 4^x-3\cdot 2^x=4 [/imath] after that substitution for [imath] y? [/imath]