Solve the equation, log4 ^(m+ 2) - log4^ (m- 5) = log4^ 8

Is there a solution to this at all?

Printable View

- Jul 9th 2007, 06:14 PMBlueStarLogarithmic Equation
Solve the equation, log4 ^(

*m*+ 2) - log4^ (*m*- 5) = log4^ 8

Is there a solution to this at all? - Jul 9th 2007, 06:20 PMJhevon
no there isn't

i assume the 4 is not the base. in that case

$\displaystyle \log 4^{m + 2} - \log 4^{m - 5} = \log 4^8$

$\displaystyle \Rightarrow \log \left( \frac {4^{m + 2}}{4^{m - 5}} \right) = \log 4^8$

$\displaystyle \Rightarrow \log 4^7 = \log 4^8$

which is not true - Jul 9th 2007, 06:59 PMJhevon
Here's another way to see that something is off with this equation

$\displaystyle \log 4^{m + 2} - \log 4^{m - 5} = \log 4^8$

$\displaystyle \Rightarrow (m + 2) \log 4 - (m - 5) \log 4 = 8 \log 4$

$\displaystyle \Rightarrow \left[ (m + 2) - (m - 5) \right] \log 4 = 8 \log 4$

equating coefficients we find

$\displaystyle \Rightarrow (m + 2) - (m - 5) = 8$

$\displaystyle \Rightarrow 7 = 8$ ---------> False - Jul 9th 2007, 07:00 PMSoroban
Hello, BlueStar!

I bet those 4's are bases . . .

Quote:

Solve the equation: .$\displaystyle \log_4(m + 2) - \log_4(m - 5) \;= \;\log_4(8)$

We have: .$\displaystyle \log_4\left(\frac{m+2}{m-5}\right) \;=\;\log_4(8)$

Then: .$\displaystyle \frac{m+2}{m-5} \;=\;8\quad\Rightarrow\quad m + 2 \;=\;8m - 40\quad\Rightarrow\quad -7m \;=\;-42$

. . Therefore: .$\displaystyle m \,=\,6$

- Jul 9th 2007, 07:01 PMJhevon
yes, i was thinking that as well. don't you just hate those typos?

i made the statement that i assumed 4 was not the base, but BlueStar said nothing about it, unless he/she doesn't know what that means ...or it really wasn't the base, let's give BlueStar the benefit of the doubt