Math Logs

[Arty]

Long Story Short i require help

Express y in terms of x

 

 

log y = 2log x

[Guest]

log y = logx^2

y = x^2

[_ASSASSIN_]

jesus at least put this in off topic

y = 10^(2logx)

y=x^2

[Arty]

log y = logx^2

y = x^2

oh so you just divide log by both sides then? 

[Guest]

oh so you just divide log by both sides then? 

If their bases are same than yes.

[_ASSASSIN_]

oh so you just divide log by both sides then? 

haha you might want to pay attention in class next time

http://www.mathsisfun.com/algebra/logarithms.html

[Arty]

haha you might want to pay attention in class next time

http://www.mathsisfun.com/algebra/logarithms.html

she didn't teach us

she likes to teach us after we get the assignment so dumb

[T.Vengeance]

oh so you just divide log by both sides then?

If their bases are same than yes.

 

A more proper way is to raise e to both sides. so e^log y = e^ log(x^2)

so then y = x^2

[Guest]

 

A more proper way is to raise e to both sides. so e^log y = e^ log(x^2)

so then y = x^2

It could be done but I dont see why is that more proper way

[_ASSASSIN_]

 A more proper way is to raise e to both sides. so e^log y = e^ log(x^2)

so then y = x^2

that would be implying that log is log base e. WHen I'm learning log is log base 10.

[T.Vengeance]

It could be done but I dont see why is that more proper way

Cause you cant divide a log(unless you want it to be 1 lol). It's a function. So to undo a function, you inverse function that function. Aka use e^ log

[T.Vengeance]

that would be implying that log is log base e. WHen I'm learning log is log base 10.

OH WHOOPS. Haha too much university math. @Arty yeah it's 10^log(whatever your inside is).

[Arty]

Cause you cant divide a log(unless you want it to be 1 lol). It's a function. So to undo a function, you inverse function that function. Aka use e^ log

My teacher will probably teach it this way tomorrow , college class in high school

[Glenwing]

oh so you just divide log by both sides then?

Log is an operation. "Dividing by log" is like saying "x^2 = y^2, just divide by squared on both sides and you get x = y!"

While the answer is correct (assuming all positive numbers) and it's correct that you solve by getting rid of the exponents, you do it by square rooting both sides which will cancel the exponents.

Similarly here, you don't "divide by log". You need to cancel them out by performing the inverse operation, in this case exponentiation.

"Log" has base 10 unless specifically stated otherwise, so "log(x)" means "the number that you raise 10 to which will give you X".

So if log(y) = log(x^2) then algebra allows you to exponentiate both sides on 10 to cancel out the log. 10^(log(y)) effectively means "10 raised to (the number that gives you Y when you raise 10 to it)" which obviously just gives you Y by definition. The same for the x^2 side, since you have to raise both sides to 10. So you're just left with y = x^2.

Course you also have to know Log rules so that you can rewrite 2log(x) as log(x^2).