You probably meant the right thing (the follow-up text indicates so). Nevertheless I'd like to point out that the correct statement would be "... OR ...", since in mathematics saying "and" implies that both conditions have to be met (simultaneuously), whereas you (hopefully) meant that x must meet at least one of the conditions.antinerd said:For the first one, tell me if this is right:
e^x(x^2-1)=0 ->
e^x = 0 and x^2 - 1 = 0
I assume you get x=1 from x²-1=0. Are you sure that is the only value x which makes that condition true?so x = 1 and 0? but 0 doesn't work when I plug it back in. so is 1 the solution for x?
Timo said:You probably meant the right thing (the follow-up text indicates so). Nevertheless I'd like to point out that the correct statement would be "... OR ...", since in mathematics saying "and" implies that both conditions have to be met (simultaneuously), whereas you (hopefully) meant that x must meet at least one of the conditions.
Timo said:I assume you get x=1 from x²-1=0. Are you sure that is the only value x which makes that condition true?
I can only guess where you got x=0 from. It doesn't really matter since x=0 does not satisfy any of the two conditions: [tex] e^0 = 1 \neq 0, \ 0^2 -1 = -1 \neq 0 [/tex].
antinerd said:Well, x can then equal 1 and -1, if x²-1=0.
Am I at least doing this part correctly, deriving the right answer? Was I correct to factor out the e^x? Because then I'm left with:
e^x(x²-1)=0 and then can't I get:
e^x = 0 or x²-1 = 0?
[tex]\ln_e{e^x}_ = \ln{0}[/tex]Then how can I figure out e^x = 0? Is there a way?
And for the second one, this is what I tried doing:
2e^(2+x)=12 <- this was the proper question, typo in the first post
then...
e^(2+x) = 6
(2+x) ln e = ln 6
(2+x) = ln 6
x = ln6 - 2
So does x = ln6 - 2?
It is true that x^2 - 1 = 0 if and only if (x = 1 or x = -1).antinerd said:Well, x can then equal 1 and -1, if x²-1=0.
Hurkyl said:It is true that x^2 - 1 = 0 if and only if (x = 1 or x = -1).
x, of course, cannot be both 1 and -1.
If you know that x^2 - 1 = 0, you are still not guaranteed that x = 1 is a solution to whatever problelm you are trying to solve. Similarly, you are not guaranteed that x = -1 is a solution. You are simply guaranteed that one of them is a solution. If you aren't even sure that x^2 - 1 = 0 is the answer -- e.g. you are only considering it as a possible constraint -- then it still might be the case that neither x = 1 nor x = -1 are solutions to your problem.
antinerd said:How come it can't be both?
x^2 e^x - e^x = 0
If I plug in either 1 or -1, it still gives me 0 as a solution, does it not?
1^2 e^1 - e^1 = 0
e^1 - e^1 = 0 <- right?
and also the same with -1:
-1^2 e^-1 - e^-1 = 0
e^-1 - e^-1 = 0 <- right?
Feldoh said:He's saying it's a constraint because when dealing with "e and natural logs" for some numbers they are undefined. It's always a good idea to plug in your answers into the original equation to make sure they are truly answers. In this example the answer is 1 or -1 since both work in the original equation (x^2)(e^x) - e^x = 0. Does that make sense?
antinerd said:Can I say for an answer:
x equals 1 AND -1
??
antinerd said:Can I say for an answer:
x equals 1 AND -1
??
Hurkyl said:If x = 1 AND x = -1, then it would follow that 1 = -1.
Plain english is slightly ambiguous here -- the phrase
x = 1 and x = -1 is a solution to x^2 - 1 = 0
has two interpretations. The following is correct:
x = 1 is a solution to x^2 - 1 = 0, and x = -1 is a solution to x^2 - 1 = 0.
The following is incorrect:
(x = 1 and x = -1) is a solution to x^2 - 1 = 0.
Well, if the author is careful about his grammar, you can tell the difference between the two; the former would be written "are solutions", not "is a solution".
Technically speaking, (x = 1 and -1) isn't even a grammatically correct phrase!
I admit that I'm being somewhat pedantic here -- but I think that's important. Experts can get away with being sloppy, but sloppiness tends to confuse students.
To solve this equation, you need to use the algebraic properties of exponents and the distributive property. First, factor out e^x from both terms: e^x * (x^2 - 1) = 0. Then, you can set each factor to equal 0 and solve for x. So, e^x = 0 or x^2 - 1 = 0. However, e^x cannot equal 0, so the only solution is x = 1 or -1.
To solve this equation, you can use the same method as the previous question. First, divide both sides by 2: e^(2+x) = 3. Then, take the natural logarithm of both sides to cancel out the e: ln(e^(2+x)) = ln(3). This leaves you with 2+x = ln(3). Finally, subtract 2 from both sides to get the solution x = ln(3) - 2.
Yes, you can use a calculator to solve these equations. However, it is important to understand the algebraic properties and the steps involved in solving the equations by hand before relying on a calculator.
Yes, there are other methods for solving equations involving exponents, such as using logarithms or graphing the equations. However, the most common and efficient method is using algebraic properties and solving for x.
To check your solution, you can plug the value of x back into the original equation and see if it satisfies the equation. For example, for the first equation (x^2)(e^x) - e^x = 0, if you plug in x = 1, you get (1^2)(e^1) - e^1 = 0, which is a true statement. This confirms that x = 1 is a valid solution for the equation.