Solve the equation, log4 ^(m + 2) - log4^ (m - 5) = log4^ 8
Is there a solution to this at all?
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
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
Hello, BlueStar!
I bet those 4's are bases . . .
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: .$\displaystyle m \,=\,6$
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