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Exponential equation

  • Thread starter Scrythe
  • Start date
  • #1
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Homework Statement


Solve the exponential equation.
1. (22x)+(2x) - 12 = 0
2. 3 * 4x + 4 * 2x + 8 = 0

Homework Equations





The Attempt at a Solution


1. 2x ln 2 + x ln 2 = ln 12
x(2 ln 2 + ln 2)= ln 12
x=( ln 12) / (2 ln 4).
I'm thinking i'm factoring out the X wrong.

2. 12 ( 4x * 2x = -8
4x * 2x = -2/3
x (ln 8 ) = -2/3
x = (-2/3) / (ln 8)
The back of the book says there is no real solution. Why? Lets say it was a real solution. Did i solve the problem correctly?

Thanks for the help.
-Scrythe
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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You are doing some unorthodox, i.e. wrong, algebra there. It's not true that ln(2^(2x)+2^x)=ln(2^(2x))+ln(2^x) or that 3*4^x+4*2^x=12*(4^x*2^x). The way to solve the first one is to notice that if you put u=2^x then 2^(2x)=u^2. So the equation turns into a quadratic u^2+u-12=0. Solve the quadratic and then figure out x from u. Can you turn the second one into a quadratic as well? Oh, and practice your algebra!!!
 
  • #3
33,086
4,793
You have algebra mistakes on almost every single line of your work.
For problem 1, you have
2x ln 2 + x ln 2 = ln 12
x(2 ln 2 + ln 2)= ln 12
x=( ln 12) / (2 ln 4).
As Dick already commented, your first equation is incorrect. The second equation has no error in it. The third equation has an error, because 2 ln 2 + ln 2 != 2 ln 4.
2 ln 2 + ln 2 = 3 ln 2 = ln 2^3 = ln 8.

Your work in problem 2 also has some errors.
12 ( 4x * 2x = -8
4x * 2x = -2/3
x (ln 8 ) = -2/3
x = (-2/3) / (ln 8)
Dick already pointed out the error in your first equation above. Going from your first equation to the next, no errors. In the third equation, the error is that 4x * 2x = 8x^2, which is not at all equal to x ln 8.

Your basic algebra skills are killing your ability to work with more complicated concepts, such as logarithms and exponents. For each of the mistakes that Dick and I pointed out, you should go back and review those topics to make sure that you really understand them. If you don't do that, you're going to have a very tough time with the current topics you're studying.


equivalent to the first equation, so no mistake.
The first line above
 
  • #4
6
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it was (4^x) * (2^x) not 4x * 2x, but nevertheless i was able to find that there was no solution because the discriminant was negative.
(12^1+x) + (8^1+x)
(u^2) + u + 8 = 0
(b^2) - 4ac = 1 - 32 = no real solution.

i was also able to solve the first one. ( though i won't post my work). I realize i am no expert in algebra, but i have not had troubles until this section. I will look back at the algebra techniques.

thanks again,
Scrythe
 
  • #5
33,086
4,793
Sorry, I just copied your work, and the superscripts got lost in the copy.

For the second problem, your answer is correct, but your work isn't. The original equation is 3*4x + 4*2x + 8 = 0
If you make the substitution u = 2x, the equation becomes 3u2 + 4u + 8 = 0. This is quite different from what you posted, which was u2 + u + 8 = 0.

I have no idea of where you got this:
(12^1+x) + (8^1+x)
 
  • #6
6
0
Ok that makes sense. I did 3 * 4^x = 12 ^(1+x) and 4 * 2^x = 8^(1+x), which i now realize is incorrect.
 
  • #7
33,086
4,793
And that's what we mean by going back over those old topics. As you go further with mathematics, if you don't have the basic stuff down cold, you'll spend all of your time making mechanical errors and won't have a prayer at understanding the higher-level concepts.
 
  • #8
365
0
And why don't you substitute y=2x and solve the quadratic equation?
 

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