MHB Help with Exponential Equation

[squexy]

Could you help me with this questions as well?

View attachment 3751This is what I have done:
2^x - 2^-x/ 2^x +2^-x
2^x - 1/2^x/ 2^x +1/2^x
u -1/u/u+1/u =1/3

3u² - 3/ 3u² +3= U

u² - 1- u = 0
u² - u - 1 =0
(u -1) (u+1) =0
U =1 or -1

 

[MarkFL]

I have moved this post to its own thread, as we ask that new questions not be tagged onto existing threads so that our threads do not become convoluted and hard to follow.

We are given to solve:

$$\frac{2^x-2^{-x}}{2^x+2^{-x}}=\frac{1}{3}$$

My first step would be to multiply the left side by:

$$1=\frac{2^x}{2^x}$$

so that the equation becomes:

$$\frac{2^{2x}-1}{2^{2x}+1}=\frac{1}{3}$$

Now, cross-multiply, and solve for $2^{2x}$, and express the other side of the equation as a power of 2 and equate the exponents. What do you find?

 

[evinda]

Could you help me with this questions as well?

https://www.physicsforums.com/attachments/3751This is what I have done:
2^x - 2^-x/ 2^x +2^-x
2^x - 1/2^x/ 2^x +1/2^x
u -1/u/u+1/u =1/3

3u² - 3/ 3u² +3= U

u² - 1- u = 0
u² - u - 1 =0
(u -1) (u+1) =0
U =1 or -1

(Wave)$$\frac{2^x(2^x-2^{-x})}{2^x(2^x+2^{-x})}=\frac{1}{3} \Rightarrow \frac{4^x-1}{4^x+1}=\frac{1}{3} \Rightarrow 3(4^x-1)=4x+1 \Rightarrow 3 \cdot 4^x-4x=1+3 \Rightarrow 2 \cdot 4^x=4 \Rightarrow 4^x=2 \Rightarrow 2^{2x}=2 \Rightarrow \log_2{2^{2x}}=\log_2 2 \Rightarrow 2x=1 \Rightarrow x=\frac{1}{2}$$

 

[SuperSonic4]

Could you help me with this questions as well?

https://www.physicsforums.com/attachments/3751This is what I have done:
$$\dfrac{2^x - 2^{-x}}{ 2^x +2^{-x}}$$

$$\dfrac{2^x - \frac{1}{2^x}}{2^x +\frac{1}{2^x}}$$[math]\dfrac{u -\frac{1}{u}}{u+\frac{1}{u}} = \dfrac{1}{3}[/math]

$$3u² - 3/ 3u² +3= U$$$$u² - 1- u = 0$$
$$u² - u - 1 =0$$
$$(u -1) (u+1) =0$$

this is not the correct factorisation, if you expand (u-1)(u+1) you get u^2-1[/color]

U =1 or -1

Assuming $$u = 2^x$$

[math]\dfrac{u -\frac{1}{u}}{u+\frac{1}{u}} = \dfrac{1}{3}[/math]

Multiply both sides by [math]u + \frac{1}{u}[/math]

[math]u - \frac{1}{u} = \dfrac{1}{3}\left(u + \frac{1}{u}\right)[/math]

Multiply both sides by [math]u[/math] (and since [math]u > 0[/math] we needn't worry about losing solutions)

[math]u^2 - 1 = \dfrac{1}{3}(u^2 + 1)[/math]

Multiply both sides by 3

[math]3u^2-3 = u^2+1[/math]

From here you should be able to solve for [math]u[/math]

---

From what I can gather on the RHS you've done

[math]\dfrac{1}{3}(u^2+1) = \dfrac{u^2}{3} + \dfrac{1}{3}[/math] which is allowed and then divided by it which would end up with

[math]\frac{3u^2-3}{\dfrac{u^2}{3} + \dfrac{1}{3}}[/math] which is not the same as you have.$$\frac{3u^2 - 3}{3u^2 +3} = U$$

 

[squexy]

My bad for posting wrong.

Thank you all for the answers.

Assuming $$u = 2^x$$

[math]\dfrac{u -\frac{1}{u}}{u+\frac{1}{u}} = \dfrac{1}{3}[/math]

Multiply both sides by [math]u + \frac{1}{u}[/math]

[math]u - \frac{1}{u} = \dfrac{1}{3}\left(u + \frac{1}{u}\right)[/math]

Multiply both sides by [math]u[/math] (and since [math]u > 0[/math] we needn't worry about losing solutions)

[math]u^2 - 1 = \dfrac{1}{3}(u^2 + 1)[/math]

Multiply both sides by 3

[math]3u^2-3 = u^2+1[/math]

From here you should be able to solve for [math]u[/math]

---

From what I can gather on the RHS you've done

[math]\dfrac{1}{3}(u^2+1) = \dfrac{u^2}{3} + \dfrac{1}{3}[/math] which is allowed and then divided by it which would end up with

[math]\frac{3u^2-3}{\dfrac{u^2}{3} + \dfrac{1}{3}}[/math] which is not the same as you have.$$\frac{3u^2 - 3}{3u^2 +3} = U$$

Is this correct?
3u² - 3 = u² + 1
2u² = 4
u = √ 2

2^x = 2^1/2
x = 1/2

 

[evinda]

My bad for posting wrong.

Thank you all for the answers.
Is this correct?
3u² - 3 = u² + 1
2u² = 4
u = √ 2

2^x = 2^1/2
x = 1/2

It is right (Nod)

But, notice that:

$$u^2=2 \Rightarrow u=\pm \sqrt{2}$$

But since $u=2^x>0$, we reject $u=-\sqrt{2}$.

 

[squexy]

It is right (Nod)

But, notice that:

$$u^2=2 \Rightarrow u=\pm \sqrt{2}$$

But since $u=2^x>0$, we reject $u=-\sqrt{2}$.

If 2^x were smaller than 0 then -√2 would be accepted and √2 woudn´t?

 

[evinda]

If 2^x were smaller than 0 then -√2 would be accepted and √2 woudn´t?

It cannot be that $2^x$ is negative, since it is an exponential function. (Shake)

But, if we would have set for example $u=-2^x$ and we would have $u^2=2$ then we would get $u=\pm \sqrt{2}$ and we would reject the case $u=+\sqrt{2}$, since in this case $u$ is always negative. (Smile)