Express y as a function of x. The constant C is a positive number. y= In 4x + In C (a) y = 4Cx (b) y = 4x + C (c) y = (4x)^C (d) y = x + 4C Need a walk through plz! :)
y is a function of x, you have a typo
can you take a screenshot of your worksheet
y IS a function of x , dude
good luck
no, you didnt copy it correctly
none of your solutions have ln x in it, notice that?
i have a math degree
how do you know im not in college? lol
you better delete that
check your parenthesees or something. the answers dont match the question
am i?
if y= In 4x + In C y does not equal 4cX
well we can get a second opinion if you dont trust me
ok bump it
you can graph it if you like
ln just doesnt disappear
What @perl means is that y is already a function of x. If y appears alone on one side of the equation, with x on the other side, then y is a function of x. :)
terenz , his answers dont match the question.
e.coiuba lets move on. why live in the past
Oh... right, are you sure there are no logarithms (ln) in your choices, @e.cociuba ?
i already posted the solution, but ill do it again -sighs heavily- y= In 4x + In C = ln (4xC) , there we are done
i did leave already, but i keep getting notifications
im not sure how to unnotify myself
My fault... @perl I'm going to stay out of this, until further notice or instruction...
terenz, wait
terenz, we need your second opinion
Okay... @e.cociuba You're not going to love this... (probably) but I agree with @perl at almost every point~ y already is expressed as a function of x and ln does not simply disappear... So, if I must have a different opinion... I'm wondering (wildly)... could there have been an expression \(\large e^y\) by any chance?
Well, trust @perl The answers are quite accurate and well-explained :)
so we are using log rules here ln (x ) + ln(y) = ln ( x*y) in this case we have ln(4x) + ln (C) = ln ( 4x*C)
None of the answer choices YOU posted are correct. But if there was in fact, a "ln" preceding every choice, then yes, a.) would be correct.
Looking at it right now, methinks :)
Ok nvm boys. Thanks for yalls for ur help :)
So many deleted comments o.O No problem @e.cociuba
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