# graphing logarithmic equation

• April 5th 2010, 01:45 AM
Anemori
graphing logarithmic equation
• April 5th 2010, 02:55 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}}}}) =
$

$= {\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 -
$

$- {\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}} =
$

$= ({\log _n}m + 1)x + ( - 1 - \frac{1}{{{{\log }_n}m}})
$

This is line (f(x)=ax+b). Can you graph this now?
• April 5th 2010, 06:54 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, 04:31 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, 04:37 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 7th 2010, 12:51 AM
Anemori
Ok the original problem is how do you graph this and finding intercepts?
• April 7th 2010, 01:20 AM