Help with Exponential Equation

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In summary, the conversation is about solving a mathematical equation and finding the value of x. The equation is manipulated by multiplying both sides by a suitable term, and then simplifying to get a quadratic equation. The solutions for u are found to be u = √2 or u = -√2, but since u = 2^x, the solution for x is x = 1/2. It is also mentioned that the value of u cannot be negative, as it is an exponential function.
  • #1
squexy
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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
 

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  • #2
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:

\(\displaystyle \frac{2^x-2^{-x}}{2^x+2^{-x}}=\frac{1}{3}\)

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

\(\displaystyle 1=\frac{2^x}{2^x}\)

so that the equation becomes:

\(\displaystyle \frac{2^{2x}-1}{2^{2x}+1}=\frac{1}{3}\)

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?
 
  • #3
squexy said:
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)$$\frac{2^x(2^x-2^{-x})}{2^x(2^x+2^{-x})}=\frac{1}{3} \Rightarrow \frac{4^x-1}{4^x+1}=\frac{1}{3} \Rightarrow 3(4^x-1)=4x+1 \Rightarrow 3 \cdot 4^x-4x=1+3 \Rightarrow 2 \cdot 4^x=4 \Rightarrow 4^x=2 \Rightarrow 2^{2x}=2 \Rightarrow \log_2{2^{2x}}=\log_2 2 \Rightarrow 2x=1 \Rightarrow x=\frac{1}{2}$$
 
  • #4
squexy said:
Could you help me with this questions as well?

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

\(\displaystyle \dfrac{2^x - \frac{1}{2^x}}{2^x +\frac{1}{2^x}}\)\(\displaystyle \dfrac{u -\frac{1}{u}}{u+\frac{1}{u}} = \dfrac{1}{3}\)

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

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

U =1 or -1
Assuming \(\displaystyle u = 2^x\)

\(\displaystyle \dfrac{u -\frac{1}{u}}{u+\frac{1}{u}} = \dfrac{1}{3}\)

Multiply both sides by \(\displaystyle u + \frac{1}{u}\)

\(\displaystyle u - \frac{1}{u} = \dfrac{1}{3}\left(u + \frac{1}{u}\right)\)

Multiply both sides by \(\displaystyle u\) (and since \(\displaystyle u > 0\) we needn't worry about losing solutions)

\(\displaystyle u^2 - 1 = \dfrac{1}{3}(u^2 + 1)\)

Multiply both sides by 3

\(\displaystyle 3u^2-3 = u^2+1\)

From here you should be able to solve for \(\displaystyle u\)



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

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

\(\displaystyle \frac{3u^2-3}{\dfrac{u^2}{3} + \dfrac{1}{3}}\) which is not the same as you have.\(\displaystyle \frac{3u^2 - 3}{3u^2 +3} = U\)
 
  • #5
My bad for posting wrong.

Thank you all for the answers.

SuperSonic4 said:
Assuming \(\displaystyle u = 2^x\)

\(\displaystyle \dfrac{u -\frac{1}{u}}{u+\frac{1}{u}} = \dfrac{1}{3}\)

Multiply both sides by \(\displaystyle u + \frac{1}{u}\)

\(\displaystyle u - \frac{1}{u} = \dfrac{1}{3}\left(u + \frac{1}{u}\right)\)

Multiply both sides by \(\displaystyle u\) (and since \(\displaystyle u > 0\) we needn't worry about losing solutions)

\(\displaystyle u^2 - 1 = \dfrac{1}{3}(u^2 + 1)\)

Multiply both sides by 3

\(\displaystyle 3u^2-3 = u^2+1\)

From here you should be able to solve for \(\displaystyle u\)



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

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

\(\displaystyle \frac{3u^2-3}{\dfrac{u^2}{3} + \dfrac{1}{3}}\) which is not the same as you have.\(\displaystyle \frac{3u^2 - 3}{3u^2 +3} = U\)

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

2^x = 2^1/2
x = 1/2
 
  • #6
squexy said:
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 $u=-\sqrt{2}$.
 
  • #7
evinda said:
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?
 
  • #8
squexy said:
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)
 

FAQ: Help with Exponential Equation

What is an exponential equation?

An exponential equation is a mathematical expression that contains a variable in the exponent. It can be written in the form y = ab^x, where a and b are constants and x is the variable.

How do you solve an exponential equation?

To solve an exponential equation, you can use logarithms or take the logarithm of both sides of the equation. You can also use the properties of exponents, such as multiplying or dividing both sides by the same base, to simplify the equation.

What are some real-life applications of exponential equations?

Exponential equations are commonly used in science and finance. They can be used to model population growth, radioactive decay, compound interest, and growth or decay of investments or assets over time.

What is the difference between an exponential equation and a linear equation?

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How can I graph an exponential equation?

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