Exponents and exponential functions

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)

Homework Statement

Solve the equation:
$$5^{1-2x} = \frac{1}{5}$$

None

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:
$$1-2x=1$$
$$-2x=0$$
To get x = 0. But, that doesn't seem to be possible, since the exponent must turn out to be negative.

2)

Homework Statement

Solve the Equation:
$$8^{x^{2} - 2x}} = \frac{1}{2}$$

None

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:
$$x^2 - 2x = 1$$ $$x^2 - 2x - 1 = 0$$
But, then I'm stuck there.

3)

Homework Statement

Solve the equation:
$$e^{x^2} = (e^{3x})(\frac{1}{e^2})$$

None

The Attempt at a Solution

I'm completely clueless on this one.

Last edited:

HallsofIvy
Homework Helper
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)

Homework Statement

Solve the equation:
$$5^{1-2x} = \frac{1}{5}$$

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").

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:
$$1-2x=1$$
$$-2x=0$$
To get x = 0. But, that doesn't seem to be possible, since the exponent must turn out to be negative.
$\frac{1}{5}= 5^{-1}$ not 51

2)

Homework Statement

Solve the Equation:
$$8^{x^{2} - 2x}} = \frac{1}{2}$$

None

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:
$$x^2 - 2x = 1$$ $$x^2 - 2x - 1 = 0$$
But, then I'm stuck there.
8= 23 so $8^{x^2- 2x}= 2^{3(x^2- 2x)}$

3)

Homework Statement

Solve the equation:
$$e^{x^2} = (e^{3x})(\frac{1}{e^2})$$

None

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, $e^{3x}(\frac{1}{e^2}= (e^{3x})(e^{-x})= e^{3x-x}= e^{2x}$. Your equation is $e^{x^2}= e^{x}$ so x2= x.

All right. For the first one, I see where you went from there. I got the answer to that to be $$x=1$$.

8= 23 so $8^{x^2- 2x}= 2^{3(x^2- 2x)}$

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

Learn the "laws of exponents": (ax)y= axy (used in 2 above) and axay= ax+y.
Here, $e^{3x}(\frac{1}{e^2}= (e^{3x})(e^{-x})= e^{3x-x}= e^{2x}$. Your equation is $e^{x^2}= e^{x}$ 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
All right. For the first one, I see where you went from there. I got the answer to that to be $$x=1$$.

Ok, so with the exponents, that would be $$3(x^2 - 2x) = 1$$. Then, factoring out an x would make it: 3x(x-2) =1 .

$$\frac{1}{2}=2^{-1}$$ not 21

Dividing it would cause the equation to look like: $$x-2 = \frac{1}{3x}$$ Then $$-2 = \frac{-2}{3x}$$ then $$-6x = -2$$ then finally I get $$x = \frac{1}{3}$$. 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:
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$. 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, $e^{3x}(\frac{1}{e^2})= (e^{3x})(e^{-x})= e^{3x-x}= e^{2x}$. Your equation is $e^{x^2}= e^{x}$ so x2= x.

$$e^{3x}(\frac{1}{e^2})= (e^{3x})(e^{-2})= e^{3x-2}$$ which gives your equation $e^{x^2}=e^{3x-2}$ giving $x^2=3x-2$. This can again be solved as a quadratic equation.

HallsofIvy
Homework Helper
All right. For the first one, I see where you went from there. I got the answer to that to be $$x=1$$.

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

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 $\frac{1}{e^2}$ as [itex]\frac{1}{e^{2x}}[itex]. Check what he wrote.