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Exponents and exponential functions

  • Thread starter carbz
  • Start date
35
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Please note, there are more questions than what I'm going to give, in total I'll be giving 3. I hope to use the information with the others.

1) 1. Homework Statement
Solve the equation:
[tex]5^{1-2x} = \frac{1}{5}[/tex]


2. Homework Equations
None

3. The Attempt at a Solution
I was thinking that to make live easier, to put the entire fraction up to the first power. From there, I would use algebra in the sense of:
[tex]1-2x=1[/tex]
[tex]-2x=0[/tex]
To get x = 0. But, that doesn't seem to be possible, since the exponent must turn out to be negative.

2) 1. Homework Statement
Solve the Equation:
[tex]8^{x^{2} - 2x}} = \frac{1}{2}[/tex]


2. Homework Equations
None

3. The Attempt at a Solution
For this one, I was thinking of setting the fraction equal to 1, then factoring that out to get two answers:
[tex]x^2 - 2x = 1[/tex] [tex]x^2 - 2x - 1 = 0[/tex]
But, then I'm stuck there.

3) 1. Homework Statement
Solve the equation:
[tex]e^{x^2} = (e^{3x})(\frac{1}{e^2})[/tex]


2. Homework Equations
None


3. The Attempt at a Solution
I'm completely clueless on this one.
 
Last edited:

HallsofIvy

Science Advisor
Homework Helper
41,728
881
Please note, there are more questions than what I'm going to give, in total I'll be giving 3. I hope to use the information with the others.

1) 1. Homework Statement
Solve the equation:
[tex]5^{1-2x} = \frac{1}{5}[/tex]


2. Homework Equations
None
Actually a very relevant equation, or, more correctly, statement, would be "if ax= ay then x= y" (the exponential functions are "one-to-one").

3. The Attempt at a Solution
I was thinking that to make live easier, to put the entire fraction up to the first power. From there, I would use algebra in the sense of:
[tex]1-2x=1[/tex]
[tex]-2x=0[/tex]
To get x = 0. But, that doesn't seem to be possible, since the exponent must turn out to be negative.
[itex]\frac{1}{5}= 5^{-1}[/itex] not 51

2) 1. Homework Statement
Solve the Equation:
[tex]8^{x^{2} - 2x}} = \frac{1}{2}[/tex]


2. Homework Equations
None

3. The Attempt at a Solution
For this one, I was thinking of setting the fraction equal to 1, then factoring that out to get two answers:
[tex]x^2 - 2x = 1[/tex] [tex]x^2 - 2x - 1 = 0[/tex]
But, then I'm stuck there.
8= 23 so [itex]8^{x^2- 2x}= 2^{3(x^2- 2x)}[/itex]

3) 1. Homework Statement
Solve the equation:
[tex]e^{x^2} = (e^{3x})(\frac{1}{e^2})[/tex]


2. Homework Equations
None


3. The Attempt at a Solution
I'm completely clueless on this one.
Learn the "laws of exponents": (ax)y= axy (used in 2 above) and axay[/sup]= ax+y.
Here, [itex]e^{3x}(\frac{1}{e^2}= (e^{3x})(e^{-x})= e^{3x-x}= e^{2x}[/itex]. Your equation is [itex]e^{x^2}= e^{x}[/itex] so x2= x.
 
35
0
All right. For the first one, I see where you went from there. I got the answer to that to be [tex]x=1[/tex].

8= 23 so [itex]8^{x^2- 2x}= 2^{3(x^2- 2x)}[/itex]
Ok, so with the exponents, that would be [tex]3(x^2 - 2x) = 1[/tex]. Then, factoring out an x would make it: 3x(x-2) =1 . Dividing it would cause the equation to look like: [tex]x-2 = \frac{1}{3x}[/tex] Then [tex]-2 = \frac{-2}{3x}[/tex] then [tex]-6x = -2[/tex] then finally I get [tex]x = \frac{1}{3}[/tex]. Is that correct?


Learn the "laws of exponents": (ax)y= axy (used in 2 above) and axay= ax+y.
Here, [itex]e^{3x}(\frac{1}{e^2}= (e^{3x})(e^{-x})= e^{3x-x}= e^{2x}[/itex]. Your equation is [itex]e^{x^2}= e^{x}[/itex] so x2= x.
Ok, since that is so, the answer could only be either 0 or 1, since both when squared equal what it orginally was.
 
Last edited:

cristo

Staff Emeritus
Science Advisor
8,056
72
All right. For the first one, I see where you went from there. I got the answer to that to be [tex]x=1[/tex].



Ok, so with the exponents, that would be [tex]3(x^2 - 2x) = 1[/tex]. Then, factoring out an x would make it: 3x(x-2) =1 .
[tex]\frac{1}{2}=2^{-1}[/tex] not 21

Dividing it would cause the equation to look like: [tex]x-2 = \frac{1}{3x}[/tex] Then [tex]-2 = \frac{-2}{3x}[/tex] then [tex]-6x = -2[/tex] then finally I get [tex]x = \frac{1}{3}[/tex]. Is that correct?
This is a quadratic equation, and so will have two solutions. In solving it, you should put the equation in the familliar form, and use the quadratic formula:
[tex] x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} [/tex]. In particular, you cannot divide by x, since you do not know it's value (it could be zero.)


Ok, since that is so, the answer could only be either 0 or 1, since both when squared equal what it orginally was.
Actually, I think HallsofIvy carried a typo through the below calculation:

Learn the "laws of exponents": (ax)y= axy (used in 2 above) and axay[/sup]= ax+y.
Here, [itex]e^{3x}(\frac{1}{e^2})= (e^{3x})(e^{-x})= e^{3x-x}= e^{2x}[/itex]. Your equation is [itex]e^{x^2}= e^{x}[/itex] so x2= x.


It should read:
[tex]e^{3x}(\frac{1}{e^2})= (e^{3x})(e^{-2})= e^{3x-2}[/tex] which gives your equation [itex]e^{x^2}=e^{3x-2}[/itex] giving [itex]x^2=3x-2[/itex]. This can again be solved as a quadratic equation.
 

HallsofIvy

Science Advisor
Homework Helper
41,728
881
All right. For the first one, I see where you went from there. I got the answer to that to be [tex]x=1[/tex].



Ok, so with the exponents, that would be [tex]3(x^2 - 2x) = 1[/tex]. Then, factoring out an x would make it: 3x(x-2) =1 . Dividing it would cause the equation to look like: [tex]x-2 = \frac{1}{3x}[/tex] Then [tex]-2 = \frac{-2}{3x}[/tex] then [tex]-6x = -2[/tex] then finally I get [tex]x = \frac{1}{3}[/tex]. Is that correct?
How did you manage to go from [itex]x-2= \frac{1}{3x}[itex] to [itex]-2= \frac{-2}{3x}[/itex]? [itex]3(x^2- 2x)= 1[/itex] is a quadratic equation, as Cristo says. In "standard form" it is [itex]3x^2- 6x- 1= 0[/itex].




Ok, since that is so, the answer could only be either 0 or 1, since both when squared equal what it orginally was.
Unfortunately, as Cristo said, I misread [itex]\frac{1}{e^2}[/itex] as [itex]\frac{1}{e^{2x}}[itex]. Check what he wrote.
 

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