#Logarithmic equations
76 messages · Page 1 of 1 (latest)
Maybe I try
Thanks
I think it's best to simplify the 2nd equation
yes
But I got lost after simplifying both
Let's consider the above
𝔸dωn𝓲²s
See what I did?
In eqn 2
Mistake
huh where
Yeh
ok
No worries
So log (y^2(x-1)) = log 3^2
yes
you can solve for x in terms of y pretty easily with the second equation
𝔸dωn𝓲²s
Notice how we can substitute xy from the second equation
and then solve for y
in the first one
Yeh
Ok let me try solving it again
I think y = 3
hmm
i could've done it wrong haha
Just to check when I substitute xy it's log( 6 * y - y^2 ) = log 3
So the logs cancel out
basically
But what I'm getting is decimal for y
you forgot the square
it's 3^2
yes
put it into
log(xy) = log(6)
then you get
log(3x) = log(6)
what has x to be
2
thank you too
For what
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