Question

log[4] (x+2) - log[4] (x-1) =1

Answer 1

\[\huge\rm log_4 (x+2) \color{red}{-} \log_4 (x-1) =1\]
apply quotient property \[\huge\rm log_b x \color{red}{-} \log_b y = \log_b \frac{ x }{ y }\]
if there is negative sign you can chang3 that to divison
subtraction ---->division
addition .-----> multiplication

Answer 2

yes i understand addition becomes multiplication and so on

Answer 3

so who would you write it \[\huge\rm log_4 (x+2) \color{red}{-} \log_4 (x-1) =1\]
?

Answer 4

That's the part that is confusing me

Answer 5

both log have same base which is 4
so change \[\huge\rm log_4 (x+2) \color{red}{-} \log_4 (x-1) \]
this to division like in this example \[\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y }\]

Answer 6

\[\log[4]\frac{ x+2 }{ x-1 } = 1 ?\]

Answer 7

yep right
now change log to exponential form \[\huge\rm log = \log_{10}\]
change log to exponential form
for example \[\huge\rm log_b x = y \]\[\huge\rm b^y = x\]