1. ## graphing logarithmic equation

How do you graph this?

2. $\displaystyle lo{g_n}({(mn)^{x - {{\log }_m}n}}) = {\log _n}(\frac{{{{(mn)}^x}}}{{{{(mn)}^{{{\log }_m}n}}}}) = {\log _n}(\frac{{{m^x}{n^x}}}{{{m^{{{\log }_m}n}} \times {n^{{{\log }_m}n}}}}) =$
$\displaystyle = {\log _n}(\frac{{{m^x}{n^x}}}{{n \times {n^{{{\log }_m}n}}}}) = {\log _n}({n^{x - 1}}) + {\log _n}(\frac{{{m^x}}}{{{n^{{{\log }_m}n}}}}) = (x - 1) + x{\log _n}m -$
$\displaystyle - {\log _n}({n^{{{\log }_m}n}}) = x - 1 + x{\log _n}m - {\log _m}n = x - 1 + x{\log _n}m - \frac{1}{{{{\log }_n}m}} =$
$\displaystyle = ({\log _n}m + 1)x + ( - 1 - \frac{1}{{{{\log }_n}m}})$
This is line (f(x)=ax+b). Can you graph this now?

3. Hello, Anemori!

Slightly different approach . . . same answer.

How do you graph this?

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

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

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

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

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

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

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

4. <--- this is Icanfly answer.

and this is Soroban answer. which one is right?

5. Originally Posted by Soroban
Hello, Anemori!

Slightly different approach . . . same answer.

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

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

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

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

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

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

How can I find points and y-interecept and x-intercept on this equation?

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

7. 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