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How to factor these equations? TRY IT. NOT AS EASY AS IT SOUNDS!

nigahiga

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Aug 6, 2013
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27
1 + 3^(x/2) = 2^x

Already know the answer, just not the steps leading up to the answer. How do you get to the answer?

Tad more difficult question.
( √(2-√3) )^x + ( √(2+√3) )^x= 2^x

How do you factor that and find the answer? Seriously?
 
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serds

Junior Member
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May 30, 2013
Messages
55
a=b[SUP]x[/SUP]
=> x= log[SUB]b[/SUB] a

 

nigahiga

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Aug 6, 2013
Messages
27
Sorry but...that doesn't really help...
 

HallsofIvy

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Jan 27, 2012
Messages
4,716
1 + 3^(x/2) = 2^x

Already know the answer, just not the steps leading up to the answer. How do you get to the answer?
Did you not understand serd's suggestion?

He said that "\(\displaystyle ln(x+y)= ln(x(1+\frac{y}{x})= ln(x)+ ln(1+ \frac(1+\frac{y}{x})\)". Do you not see how that is connected with this problem? Taking the logarithm of both sides of the original equation, ln(1+ 3^{x/2})= ln(2^x) which is equivalent to \(\displaystyle -ln(2^x)+ ln(1+ 3^{x/2})= ln(2^{-x}+ ln(1+ 3^{x/2})\). That matches serd's equation if his "x" is your \(\displaystyle 2^{-x}\) while his "\(\displaystyle 1+\frac{y}{x}\)" is your \(\displaystyle 1+ 3^{x/2}\). That means you must have \(\displaystyle 1+ \frac{y}{2^{-x}}= 1+ 3^{x/2}\) so that \(\displaystyle \frac{y}{2^{-x}}= 3^{x/2}\) and \(\displaystyle y= 3^{x/2}2^x= \left(2\sqrt{3}\right)^x\). That means that his ln(x+y) is your \(\displaystyle ln(2^{-x}+ \left(2\sqrt{3}\right)^x)= 0\).

Tad more difficult question.
( √(2-√3) )^x + ( √(2+√3) )^x= 2^x

How do you factor that and find the answer? Seriously?
 

nigahiga

New member
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Aug 6, 2013
Messages
27
If you know the answer, that means nothing: from back of the book?
You're evidently trying to get your homework done.
Denis, I am aware that the answer alone is not sufficient. Why else would I ask this question? I wish to learn the process of how to get to the answer, rather than just blindly plug in numbers by conjecture.

As for your second remark, this is summer. All schools are out, at least all high schools in the Southern California section vicinity. This is NOT homework, just a leisurely way to spend my time. I realize this may sound very odd and dubious, especially considering the stereotyped number of people who disregard schoolwork, but rest assured, that is honestly my sole reason for posting.

As for all else who posted, I apologize if I could not deduce the hints you guys gave me. I was imprudent and forced myself to look for a direct, more apparent answer. I will be more patient next time!
 

nigahiga

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Messages
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Did you not understand serd's suggestion?

He said that "\(\displaystyle ln(x+y)= ln(x(1+\frac{y}{x})= ln(x)+ ln(1+ \frac(1+\frac{y}{x})\)". Do you not see how that is connected with this problem? Taking the logarithm of both sides of the original equation, ln(1+ 3^{x/2})= ln(2^x) which is equivalent to \(\displaystyle -ln(2^x)+ ln(1+ 3^{x/2})= ln(2^{-x}+ ln(1+ 3^{x/2})\). That matches serd's equation if his "x" is your \(\displaystyle 2^{-x}\) while his "\(\displaystyle 1+\frac{y}{x}\)" is your \(\displaystyle 1+ 3^{x/2}\). That means you must have \(\displaystyle 1+ \frac{y}{2^{-x}}= 1+ 3^{x/2}\) so that \(\displaystyle \frac{y}{2^{-x}}= 3^{x/2}\) and \(\displaystyle y= 3^{x/2}2^x= \left(2\sqrt{3}\right)^x\). That means that his ln(x+y) is your \(\displaystyle ln(2^{-x}+ \left(2\sqrt{3}\right)^x)= 0\).
I still don't get it. Could you please elaborate? How do you use the resulting equations to get the answer?
 

HallsofIvy

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Jan 27, 2012
Messages
4,716
"The resulting answer" to what question? You titled this thread "How to factor these equations". But one does NOT factor equations, one factors expressions, possibly in order to solve an equation.

In order to solve the equation \(\displaystyle 1+ 3^{x/2}= 2^x\), First write it as \(\displaystyle 2^x- 3^{x/2}= 1\). Since that involves exponentials, take the logarithm of both sides to get \(\displaystyle ln(2^x- 3^{x/2})= ln(1)= 0\).
Now use that this is \(\displaystyle ln(2^x(1- \frac{3^{x/2}}{2^x})= ln(2^x(1- \left(\frac{3^{1/2}{2}\right)^x)= 0\)
 

HallsofIvy

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Jan 27, 2012
Messages
4,716
"The resulting answer" to what question? You titled this thread "How to factor these equations". But one does NOT factor equations, one factors expressions, possibly in order to solve an equation.

A problem like this, with the "x" in the exponent, you do not solve by factoring. You need to use the inverse function to the exponential which is the logarithm. In order to solve the equation \(\displaystyle 1+ 3^{x/2}= 2^x\), First write it as \(\displaystyle 2^x- 3^{x/2}= 1\). Since that involves exponentials, take the logarithm of both sides to get \(\displaystyle ln(2^x- 3^{x/2})= ln(1)= 0\).
Now use that this is \(\displaystyle ln(2^x(1- \frac{3^{x/2}}{2^x})=\)\(\displaystyle ln(2^x(1- \left(\frac{3^{1/2}}{2}\right)^x)= 0\)
\(\displaystyle xln(2)+ ln(1- \left(\frac{3^{1/2}}{2}\right)^x)= 0\)
 

Subhotosh Khan

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Staff member
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Jun 18, 2007
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18,086
1 + 3^(x/2) = 2^x

Already know the answer, just not the steps leading up to the answer. How do you get to the answer?

Tad more difficult question.
( √(2-√3) )^x + ( √(2+√3) )^x= 2^x

How do you factor that and find the answer? Seriously?
The only way I can get the solution for 'x' (both the cases x = 2), is to use numerical method (Newton-Raphson).

By plotting the functions, I see that x=2 is the only solution for either case.
 

stapel

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Feb 4, 2004
Messages
15,937
How do you factor that and find the answer?
On what basis had you concluded that factoring was involved? Did the instructions specify such?

Please be complete. Thank you. ;)
 

nigahiga

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Aug 6, 2013
Messages
27
"The resulting answer" to what question? You titled this thread "How to factor these equations". But one does NOT factor equations, one factors expressions, possibly in order to solve an equation.

In order to solve the equation \(\displaystyle 1+ 3^{x/2}= 2^x\), First write it as \(\displaystyle 2^x- 3^{x/2}= 1\). Since that involves exponentials, take the logarithm of both sides to get \(\displaystyle ln(2^x- 3^{x/2})= ln(1)= 0\).
Now use that this is \(\displaystyle ln(2^x(1- \frac{3^{x/2}}{2^x})= ln(2^x(1- \left(\frac{3^{1/2}{2}\right)^x)= 0\)
On what basis had you concluded that factoring was involved? Did the instructions specify such?

Please be complete. Thank you. ;)
May I ask as to why this community of experts is so condescending? I am truly apologetic for not being able to discern the correct terminology. I am not a pretender of knowledge, and I know my faults. I now realize the irrationality of my phrasing the title thread "How to factor these equations" and fully acknowledge my lack of holistic mathematics cognition. I am deeply sorry.
 

stapel

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Staff member
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Messages
15,937
May I ask as to why this community of experts is so condescending?
It is to be regretted that you view requests for clarification to constitute condescension.

In future, it might be helpful if you presented clear information (such as the full and exact text of the question, the actual instructions, a clear listing of your thoughts and efforts so far, the background of the exercise, etc) rather than hoping that the volunteers can somehow correctly guess on your behalf. Thank you. ;)
 

JeffM

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Sep 14, 2012
Messages
3,174
May I ask as to why this community of experts is so condescending? I am truly apologetic for not being able to discern the correct terminology. I am not a pretender of knowledge, and I know my faults. I now realize the irrationality of my phrasing the title thread "How to factor these equations" and fully acknowledge my lack of holistic mathematics cognition. I am deeply sorry.
Your sarcasm is misplaced. You obviously failed to read Read Before Posting. In other words, you barge in and do nothing that has been requested of you such as posting the complete exercise.

I spent considerable time (utterly wasted) trying to figure out how to solve the equations through factoring. I could not and so did not respond. That does not mean much, but Subhotosh Khan said that he was able to solve the equations only by using Newton-Raphson. That means he also could not solve them through factoring. So you wasted the time and energy of any number of people because you did not have the courtesy to ask a question in the way that we have requested. Then you get your nose out of joint because people are curt in response to your rudeness. No one is obliged to respond politely to your lack of manners.
 

nigahiga

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Aug 6, 2013
Messages
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Your sarcasm is misplaced. You obviously failed to read Read Before Posting. In other words, you barge in and do nothing that has been requested of you such as posting the complete exercise.

I spent considerable time (utterly wasted) trying to figure out how to solve the equations through factoring. I could not and so did not respond. That does not mean much, but Subhotosh Khan said that he was able to solve the equations only by using Newton-Raphson. That means he also could not solve them through factoring. So you wasted the time and energy of any number of people because you did not have the courtesy to ask a question in the way that we have requested. Then you get your nose out of joint because people are curt in response to your rudeness. No one is obliged to respond politely to your lack of manners.
I was not being sarcastic in the least. I was expressing a sincere apology for not being able to phrase the question correctly. It was to be implied that the question was to find the value of x that satisfies the equation, and it was my mistake to assume that everyone would discern the manner of the question in a unanimous manner.
 

nigahiga

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Aug 6, 2013
Messages
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Your sarcasm is misplaced. You obviously failed to read Read Before Posting. In other words, you barge in and do nothing that has been requested of you such as posting the complete exercise.

I spent considerable time (utterly wasted) trying to figure out how to solve the equations through factoring. I could not and so did not respond. That does not mean much, but Subhotosh Khan said that he was able to solve the equations only by using Newton-Raphson. That means he also could not solve them through factoring. So you wasted the time and energy of any number of people because you did not have the courtesy to ask a question in the way that we have requested. Then you get your nose out of joint because people are curt in response to your rudeness. No one is obliged to respond politely to your lack of manners.
Also, the paper in which I am getting the question from does not state any specific directions. It simply states, "solve"; in other words, what I posted was more or less the same as posting the complete exercise. Again, it was my mistake to deduce that factoring would help find the value of x, but I now realize that factoring is the wrong terminology to use in this case. I don't know how else to apologize for intruding into the forum without proper decorum! Please believe me, I really am sorry for hinting an erroneous method in evaluating the equation and taking up your time.
 

nigahiga

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Aug 6, 2013
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It is to be regretted that you view requests for clarification to constitute condescension.

In future, it might be helpful if you presented clear information (such as the full and exact text of the question, the actual instructions, a clear listing of your thoughts and efforts so far, the background of the exercise, etc) rather than hoping that the volunteers can somehow correctly guess on your behalf. Thank you. ;)
I apologize for jumping to conclusions about your requests for clarification, as I assumed it was in a belittling manner. The worksheet in which I transcribed the question from merely stated "Solve the following equations", and so I was unable to come up with a valid background of the exercise. Therefore, I erroneously used the term "factoring" interchangeably with "solving", which I now realize to be a faulty misconception. Again, sorry for any and ALL misinterpretations on my part.
 
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JeffM

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Sep 14, 2012
Messages
3,174
I apologize for jumping to conclusions about your requests for clarification, as I assumed it was in a belittling manner. The worksheet in which I transcribed the question from merely stated "Solve the following equations", and so I was unable to come up with a valid background of the exercise. Therefore, I erroneously used the term "factoring" interchangeably with "solving", which I now realize to be a faulty misconception. Again, sorry for any and ALL misinterpretations on my part.
OK then.

It would still help to know the context of the exercise, but Subhotosh Khan indicates 2 methods of approximate solution. You may not realize it, but most equations cannot be solved exactly and can only be solved approximately.

\(\displaystyle f(x) = g(x) \implies f(x) - g(x) = 0.\ Let\ h(x) = f(x) - g(x).\)

Now graph h(x) and note where it intersects the x-axis. If you are allowed to use graphing software, this is frequently the quickest method to get an approximate solution and sometimes it will suggest an exact solution.

The Newton-Raphson method is an iterative method for improving an initial approximation found by graphing or knowledge of the underlying problem from outside math. In many cases, Newton-Raphson gives an excellent approximation in very few iterations if you can start with a decent initial approximation. The problem for you may be that Newton-Raphson requires differential calculus, and we do not know whether you know that.

Please remember to give some background about yourself (eg high school student doing review over the summer in preparation for pre-calc) so we have an idea about how best to respond. Also please give a complete copy of the problem. There may be a clue that you are missing when you try to translate. I know the word "factoring" threw me way off.
 
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nigahiga

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OK then.

It would still help to know the context of the exercise, but Subhotosh Khan indicates 2 methods of approximate solution. You may not realize it, but most equations cannot be solved exactly and can only be solved approximately.

\(\displaystyle f(x) = g(x) \implies f(x) - g(x) = 0.\ Let\ h(x) = f(x) - g(x).\)

Now graph h(x) and note where it intersects the x-axis. If you are allowed to use graphing software, this is frequently the quickest method to get an approximate solution and sometimes it will suggest an exact solution.

The Newton-Raphson method is an iterative method for improving an initial approximation found by graphing or knowledge of the underlying problem from outside math. In many cases, Newton-Raphson gives an excellent approximation in very few iterations if you can start with a decent initial approximation. The problem for you may be that Newton-Raphson requires differential calculus, and we do not know whether you know that.

Please remember to give some background about yourself (eg high school student doing review over the summer in preparation for pre-calc) so we have an idea about how best to respond. Also please give a complete copy of the problem. There may be a clue that you are missing when you try to translate. I know the word "factoring" threw me way off.
Thank you very much! :)
I will elaborate on my background information. I am currently a high school student to be in the senior class this coming school year. I have taken first year differential calculus and am aware of the Newton-Raphson method (in my school we call simply refer to it as Newton's method). I will be moving up to second year calculus when I start school this month. Our school frequently hosts math competitions (somewhat similar to that of the AMC), and I wish to excel in this category, as very few tend to do well in my school. The questions I have asked and the questions TO be asked is a little portion of the questions the competitions showcase. Very few prepare for these competitions, most likely because most of my high school peers are not interested in mathematics as an intriguing subject and simply take the AP courses to get credit for colleges. I find these questions engaging and wish to actually prepare for these competitions. It may sound weird coming from a high school student whose introduction to this forum was rash to say the least, but that is the truth of the matter. This is NOT for mandatory homework or a summer study task as someone previously mentioned!
 

nigahiga

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Aug 6, 2013
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I have attached a copy of some of the other questions at hand. It's a little messy, as I have taken each question from different worksheets and combined them into one page. I will now copy the directions onto this post exactly as stated on the questions' respective worksheets.
#1. Solve for x.
#2. Simplify.
#4. Solve for x and y.
#6. Solve for x.
#9 Solve for x and y.
#13. Simplify.
#19. Solve for x.
#3. Prove the trigonometric identity. (Direction doesn't state restrictions such as "only alter the left side", so do as you will)

I don't need all of these to be solved (or any, for that matter). I want to see your thoughts on these questions! And if you would so desire, shed some light on how to solve these problems! I have utilized WolframAlpha to find the solutions, yet I do not know how to get there.
Again, I apologize for the messiness. It is very difficult to organize these questions because of the sheer quantity of questions!

EDIT: One more note! The administrators of these competitions do not allow for graphing utilities or computer-oriented calculators. Scientific calculators are the most that they will allow. Therefore, they state that all questions administered are possible without the use of overly-advanced mathematics utilities. For example, all "Solve for x" questions are possible without graphing the equations to determine the zeroes of the function. Be that as it may, I am still not sure as to how my initial question (√(2+√3))^x + (√(2-√3))^x = 2^x can be solved without a graphing utility, as the proofs and answer keys to these questions are locked away in a mathematical vault (figurative, but true). So if you guys can find other methods of determining the solutions, that would be awesome!
2ND EDIT: Also, the competitions limit the ways of solving these questions to Pre-Calculus methods (so no Newton's Method, differential calculus).
 

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Dale10101

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Feb 25, 2013
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318
gracias

1 + 3^(x/2) = 2^x

Already know the answer, just not the steps leading up to the answer. How do you get to the answer?

Tad more difficult question.
( √(2-√3) )^x + ( √(2+√3) )^x= 2^x

How do you factor that and find the answer? Seriously?
For me, a fruitful question, the ensuing dialog illuminated:

1) The difference between an expression and an equation.

(How long did it take me to began to realize that there is a hierarchy in algebra of beginning with the definition of fundamental operators (+,-,*,"/", exponentiation) followed by the construction of expressions as an amalgam of fundamental operators, followed by the use of the equal sign in one of two ways, a) equating two expressions, b) assigning a name to an expression to define a function.)

2) The realization that factoring is a procedure applied to expressions not equations which emphasizes the importance of distinguishing between the two.

(Regarding factoring: I only recently realized the Fundamental law of Algebra is a sort of analog of the fundamental law of Arithmetic. That is, that just as integers can always be reduced to a product of prime numbers, so to polynomials can always be reduced to a product of linear factors if you factor of the field of complex numbers.)

BTW. I am not a mathematician so if my insights are limited or erroneous I am happy for correction, nor am I going to embarrassed by the fact that such insights are merely pedagogic.

3) That not all equations can analytically be solved, i.e. transposed to a functional form.

An insight here is that an equation is, in a sense fundamentally a yes/no proposition. That is, an equation defines a subset of the Cartesian product of its variable set, so, in an impractical sort of way you could find this solution subset by feeding in each element of the said Cartesian product for a yes/not, accept/reject response.

Impracticable naturally, “Simple pimple” of course, but for me, large scale concepts like that allow me not freak out when confronted by something new … to at least know what ball park I am operating in.

Your question allowed me to learn something new and to confirm something old, so … thanks.
 
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