1. ## Logarithmic Equation

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

Is there a solution to this at all?

2. Originally Posted by BlueStar
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

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

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

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

which is not true

3. Originally Posted by BlueStar
Solve the equation, log4 ^(m + 2) - log4^ (m - 5) = log4^ 8

Is there a solution to this at all?
Here's another way to see that something is off with this equation

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

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

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

equating coefficients we find

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

$\Rightarrow 7 = 8$ ---------> False

4. Hello, BlueStar!

I bet those 4's are bases . . .

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

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

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

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

5. Originally Posted by Soroban
Hello, BlueStar!

I bet those 4's are bases . . .

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

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

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

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