graphing logarithmic equation

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April 5th 2010, 12:45 AM
Anemori

graphing logarithmic equation

How do you graph this?

http://www.mathway.com/math_image.as...030?p=238?p=22
April 5th 2010, 01:55 AM
ICanFly

$lo{g_n}({(mn)^{x - {{\log }_m}n}}) = {\log _n}(\frac{{{{(mn)}^x}}}{{{{(mn)}^{{{\log }_m}n}}}}) = {\log _n}(\frac{{{m^x}{n^x}}}{{{m^{{{\log }_m}n}} \times {n^{{{\log }_m}n}}}}) = <br />$
$= {\log _n}(\frac{{{m^x}{n^x}}}{{n \times {n^{{{\log }_m}n}}}}) = {\log _n}({n^{x - 1}}) + {\log _n}(\frac{{{m^x}}}{{{n^{{{\log }_m}n}}}}) = (x - 1) + x{\log _n}m - <br />$
$- {\log _n}({n^{{{\log }_m}n}}) = x - 1 + x{\log _n}m - {\log _m}n = x - 1 + x{\log _n}m - \frac{1}{{{{\log }_n}m}} = <br />$
$= ({\log _n}m + 1)x + ( - 1 - \frac{1}{{{{\log }_n}m}})<br />$
This is line (f(x)=ax+b). Can you graph this now?
April 5th 2010, 05:54 AM
Soroban

Hello, Anemori!

Slightly different approach . . . same answer.

Quote:

How do you graph this?

. . $f(x) \;=\;\log_n(mn)^{x-\log_mn}\,\;\;m,n > 0$

$f(x) \;=\;\log_n\bigg[m^{x-\log_mn}\cdot n^{x-\log_mn}\bigg]$

. . . $=\;\log_n\left(m^{x-\log_mn}\right) + \log_n\left(n^{x-\log_mn}\right)$

. . . $=\; (x-\log_mn)\log_nm + (x-\log_mn)\underbrace{\log_nn}_{\text{This is 1}}$

. . . $=\; (\log_nm)x - \underbrace{(\log_mn)(\log_nm)}_{\text{This is 1}} + x - \log_mn$

$f(x) \;=\; \left(1 + \log_nm\right)x - \left(1 + \log_mn\right)$

We have: . $f(x) \:=\:mx + b\quad\hdots\;\text{ a line}$
April 6th 2010, 03:31 PM
Anemori

http://www.mathhelpforum.com/math-he...067f1e18-1.gif <--- this is Icanfly answer.

and this is Soroban answer. which one is right?
http://www.mathhelpforum.com/math-he...f98c9e72-1.gif
April 6th 2010, 03:37 PM
Anemori

Quote:

Originally Posted by Soroban

Hello, Anemori!

Slightly different approach . . . same answer.

$f(x) \;=\;\log_n\bigg[m^{x-\log_mn}\cdot n^{x-\log_mn}\bigg]$

. . . $=\;\log_n\left(m^{x-\log_mn}\right) + \log_n\left(n^{x-\log_mn}\right)$

. . . $=\; (x-\log_mn)\log_nm + (x-\log_mn)\underbrace{\log_nn}_{\text{This is 1}}$

. . . $=\; (\log_nm)x - \underbrace{(\log_mn)(\log_nm)}_{\text{This is 1}} + x - \log_mn$

$f(x) \;=\; \left(1 + \log_nm\right)x - \left(1 + \log_mn\right)$

We have: . $f(x) \:=\:mx + b\quad\hdots\;\text{ a line}$

How can I find points and y-interecept and x-intercept on this equation?
April 6th 2010, 11:51 PM
Anemori

Ok the original problem is how do you graph this and finding intercepts?
April 7th 2010, 12:20 AM
mathaddict

Quote:

Originally Posted by Anemori

Ok the original problem is how do you graph this and finding intercepts?

they are both correct , check the properties of logarithms .

As for the x-intercept , when f(x)=0 , x=...

y-intercept , when x=0 , f(0)=...

calculate that