Solving Inverse Plot 2log(y)=log2+log(x+1)

Thread starter hadi amiri 4
Start date Jul 11, 2008
Tags

Inverse Plot

AI Thread Summary

The equation 2log(y) = log2 + log(x+1) simplifies to y^2 = 2(x+1), which is not the inverse of the equation Y^2 = x + 2. The discussion clarifies that while logarithmic properties are applied, the resulting equation represents a different relationship. Participants emphasize the importance of correctly identifying the transformations involved. Ultimately, the conclusion is that the original equation does not yield the inverse as initially suggested.

Jul 11, 2008

hadi amiri 4

Homework Statement

what is the plot of
2log(y)=log2+log(x+1)

Homework Equations

is it the inverse of this
Y^2=x+2

The Attempt at a Solution

Physics news on Phys.org

Jul 11, 2008

rock.freak667

Homework Helper

Homework Equations

log_a x^n=nlog_a x

log_a x + log_a y= log_a xy

Use those two equations and you'll see.

Jul 13, 2008

hadi amiri 4

is inverse of y^2=2(x+1)

Jul 13, 2008

HallsofIvy

Science Advisor

Homework Helper

No, it not the inverse. What you have is exactly y²= 2(x+1)

Last edited by a moderator: Jul 13, 2008

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

View full post »