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

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

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?

2. a=bx
=> x= logb a

3. Sorry but...that doesn't really help...

4. Originally Posted by nigahiga
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 "$ln(x+y)= ln(x(1+\frac{y}{x})= ln(x)+ ln(1+ \frac(1+\frac{y}{x})$". 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 $-ln(2^x)+ ln(1+ 3^{x/2})= ln(2^{-x}+ ln(1+ 3^{x/2})$. That matches serd's equation if his "x" is your $2^{-x}$ while his "$1+\frac{y}{x}$" is your $1+ 3^{x/2}$. That means you must have $1+ \frac{y}{2^{-x}}= 1+ 3^{x/2}$ so that $\frac{y}{2^{-x}}= 3^{x/2}$ and $y= 3^{x/2}2^x= \left(2\sqrt{3}\right)^x$. That means that his ln(x+y) is your $ln(2^{-x}+ \left(2\sqrt{3}\right)^x)= 0$.

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

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

5. Originally Posted by nigahiga
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?
If you know the answer, that means nothing: from back of the book?
You're evidently trying to get your homework done.

In this financial formula:
(1 + i)^n = F
can you solve for n?

6. Originally Posted by Denis
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!

7. Originally Posted by HallsofIvy
Did you not understand serd's suggestion?

He said that "$ln(x+y)= ln(x(1+\frac{y}{x})= ln(x)+ ln(1+ \frac(1+\frac{y}{x})$". 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 $-ln(2^x)+ ln(1+ 3^{x/2})= ln(2^{-x}+ ln(1+ 3^{x/2})$. That matches serd's equation if his "x" is your $2^{-x}$ while his "$1+\frac{y}{x}$" is your $1+ 3^{x/2}$. That means you must have $1+ \frac{y}{2^{-x}}= 1+ 3^{x/2}$ so that $\frac{y}{2^{-x}}= 3^{x/2}$ and $y= 3^{x/2}2^x= \left(2\sqrt{3}\right)^x$. That means that his ln(x+y) is your $ln(2^{-x}+ \left(2\sqrt{3}\right)^x)= 0$.
I still don't get it. Could you please elaborate? How do you use the resulting equations to get the answer?

8. "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 $1+ 3^{x/2}= 2^x$, First write it as $2^x- 3^{x/2}= 1$. Since that involves exponentials, take the logarithm of both sides to get $ln(2^x- 3^{x/2})= ln(1)= 0$.
Now use that this is $ln(2^x(1- \frac{3^{x/2}}{2^x})= ln(2^x(1- \left(\frac{3^{1/2}{2}\right)^x)= 0$

9. "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 $1+ 3^{x/2}= 2^x$, First write it as $2^x- 3^{x/2}= 1$. Since that involves exponentials, take the logarithm of both sides to get $ln(2^x- 3^{x/2})= ln(1)= 0$.
Now use that this is $ln(2^x(1- \frac{3^{x/2}}{2^x})=$$ln(2^x(1- \left(\frac{3^{1/2}}{2}\right)^x)= 0$
$xln(2)+ ln(1- \left(\frac{3^{1/2}}{2}\right)^x)= 0$

10. Originally Posted by nigahiga
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.

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