log subscript 6 x + log subscript 6 (x-5)=2
your lesson is really funny =)) \[\log_6 x + \log_6 (x-5) = 2\] do you know laws of logarithms
no i actually never took this math before thats the problem
the product law of logarithms states that: \[\Large \log_a b + \log_a c = \log_a (bc)\] do you get the application?
i mean how you can apply the law on this problem?
by getting the logs of the two 6's and combining them thus creating (bc ?? 0_0
yup..the question is..what is bc in this case?
0.778?? and for the x-5 would u use the natural log
what didyou do?
i took the log of 6?
hmm not quite..
let's try this again \[\LARGE \log_6 x + \log_6 (x+5)\] doyou agree that this is in the form \[\huge \log_a b + \log_a c\]
yes
what's a in \[\large \log_6 x + \log_6 (x+5)\]
x
nope look carefullywhere a is placed in \[\huge \log_a b + \log_a c\]
6?
yes 6. what is b?
x
and c?
0?
nope.. look carefully at \[\huge \log_6 x + \log_6 (x+5)\] and compare it with \[\huge \log_a b + \log_a c\] hint: we already decided a = 6 and b = x
5
nope.. let's try to substitute a and b into 6 and x respectively \[\huge \log_a b + \log_a c = \log_6 x + \log_6 c\] what is that c supposed to be?
:( idk
notice the difference here...\ \[\Huge \log_6 x + \log_6 c \rightarrow \log_6 x + \log_6 (x-5)\]
what can yo say?
idk
what do you see as difference?
spot the difference..
If the base are same then the following property is used; i.e\[\log_{a}{b}+\log_{a}{c}=\log_{a}{bc} \] now, use this property to the question as follows \[\log_{6}{x} +\log_{6}{x-5}=2=>\log_{6}{x^2-5x}=2 \] now change it into exponential form as\[6^2=x^2-5x\] \[=>36=x^2-5x]/ \[=>x^2-5x-36=0\] now solve this quadratic equation & get ur answer.
understood @sharee2012
no
then, why did u give me medal???
-_-
-4?
Join our real-time social learning platform and learn together with your friends!