Solving a Simple Log Problem: 4^x + 4^-x = 5/2

  • Thread starter Thread starter courtrigrad
  • Start date Start date
  • Tags Tags
    Log
AI Thread Summary
The discussion revolves around solving the equation 4^x + 4^-x = 5/2. A proposed solution involves using logarithms, but the initial approach is flawed due to a misunderstanding of logarithmic properties. Participants suggest rewriting the equation by substituting y = 4^x, leading to the quadratic form y^2 + 1 = (5/2)y. This transformation clarifies the problem and allows for easier manipulation. The conversation emphasizes the importance of correctly applying algebraic principles to solve logarithmic equations.
courtrigrad
Messages
1,236
Reaction score
2
Hello,

I need help with this problem:


4^x + 4^-x = 5/2

My Solution: (Assume we use log with base 4)

log ( 4^x + 4^-x) = log(5/2)

= log (4^0) = log (5/2) ?


I don't see what I am doing wrong. I used the Product Rule. Any help would be greatly appreciated.

Thanks
 
Last edited:
Physics news on Phys.org
How about this

(4^x)(4^x) + (4^-^x)(4 ^x) = \frac{5}{2}(4^x)

Maybe i should arrange

(4^x)^2 + 1 = \frac{5}{2}(4^x)

x= 4^x

x^2 + 1 = \frac{5}{2}x
 
Last edited:
<br /> \begin{multline*}<br /> \begin{split}<br /> &amp;log(AB)=log\ A+log\ B\\<br /> &amp;There\ isn&#039;t\ anything\ like\ this:\\<br /> &amp;log(A+B)=log(AB)\\<br /> &amp;4^x+4^{-x}=4^x+\frac{1}{4^x}=\frac{4^{2x}+1}{4^x}\\<br /> &amp;4^x*4^{-x}=4^{x-x}=1\\<br /> \end{split}<br /> \end{multline*}<br />
 
Cyclovenom said:
How about this

(4^x)(4^x) + (4^-^x)(4 ^x) = \frac{5}{2}(4^x)

Maybe i should arrange

(4^x)^2 + 1 = \frac{5}{2}(4^x)

x= 4^x

x^2 + 1 = \frac{5}{2}x

x= 4^x should not be used. I see where you are going here. But this is wrong. Instead we should assign it to a different variable like y= 4^x. So,

y^2 + 1 = \frac{5}{2}y
 
It's true, i was just reminding him of Ax^2 + Bx + C, but thanks for pointing it out.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top