Quick Math Question..

[tjanuranus]

So i'm studying for my test on tuesday and i guess i'm in burn out mode because the answer i get is wrong and i understand why my teacher does it this way but i don't understand why my way doesn't work...

The question is...

3e^-x = 0.06

now what i did was divide both sides by 3. then took the ln of both sides then multiplied by negative 1 to get X.

But the answer my teacher gave is...

ln3-x = ln0.06



so what's the deal why can't i divide by 3?

[Adami]

so what's the deal why can't i divide by 3?

Because there's too many of them.

[tjanuranus]

so what's the deal why can't i divide by 3?

Because there's too many of them.

what do you mean? it's 3 Multiplied by e^-x. I can't divide the 3 out?

[Fiery Winds]

Could you clarify your teacher's answer?  Your answer is correct, btw.  I'm thinking your teacher took the natural log first.

EDIT:  Yeah, that's what your teacher did.

ln(3e^-x) = ln(0.06)   *Take natural log of both sides*

ln(3) + ln(e^-x) = ln(0.06) *Multiplication property of logarithms*

ln(3) - x = ln(0.06) *Simplify*

Both methods are correct.

[atmyne]

3e^-x = 0.06
 
 e^-x = 0.02

 x.log(e^-1) =  log(0.02)
 
           x  = log(0.02) / log(e^-1)   = 3.912023....

thats what I got lol.
Can anyone explain the difference between ln and Log? that confuses me :L
 

[Fiery Winds]

Can anyone explain the difference between ln and Log? that confuses me :L
 

ln (the natural log) is a special case logarithm with a base of e.  When not specified, a Log function has a base of 10. 

[tjanuranus]

Could you clarify your teacher's answer?  Your answer is correct, btw.  I'm thinking your teacher took the natural log first.

EDIT:  Yeah, that's what your teacher did.

ln(3e^-x) = ln(0.06)   *Take natural log of both sides*

ln(3) + ln(e^-x) = ln(0.06) *Multiplication property of logarithms*

ln(3) - x = ln(.06) *Simplify*

Both methods are correct.

WEll thanks for clarifying that my answer is correct! I was going crazy. So leaving it like he left it is correct? I get what he did with the multi properly but i thought what i did was simpler lol.

[Fiery Winds]

I'm not sure which answer he would prefer, especially because he didn't even solve for x at all. 

[atmyne]

Can anyone explain the difference between ln and Log? that confuses me :L
 

ln (the natural log) is a special case logarithm with a base of e.  When not specified, a Log function has a base of 10. 

ah yes of course. cheers. been a while since ive done a "proper" math question/class

[kári]

3e^-x = 0.06
e^-x = 0.02
ln(e^-x) = ln(0.02)
(x)*ln(e^-1) = ln(1/50)
(x)*[ln(1) - ln(e)]  = ln(1) - ln(50)
x = [ln(1) - ln(50)]/[ln(1) - 1]

... or whatever you said. You can always use www.wolframalpha.com if you're unsure of an answer... Just type the problem in. Like this:

[rumborak]

3e^-x = 0.06
e^-x = 0.02
ln(e^-x) = ln(0.02)
(x)*ln(e^-1) = ln(1/50)
(x)*[ln(1) - ln(e)]  = ln(1) - ln(50)
x = [ln(1) - ln(50)]/[ln(1) - 1]

Maybe it's just experience, but to me I go straight from ln(e^-x) to -x

rumborak

[tjanuranus]

3e^-x = 0.06
e^-x = 0.02
ln(e^-x) = ln(0.02)
(x)*ln(e^-1) = ln(1/50)
(x)*[ln(1) - ln(e)]  = ln(1) - ln(50)
x = [ln(1) - ln(50)]/[ln(1) - 1]

Maybe it's just experience, but to me I go straight from ln(e^-x) to -x

rumborak

Ok here is what my teacher taught me. LN both sides and the e cancels out. Then in the calculator push ln then .02. The of course reverse sign but yeah.. that stuff kari put up.... i dunno lol.

[Fiery Winds]

The reason it cancels out is because by performing this operation: ln(e^-x) you're asking yourself what power do you raise e to, in order to get e^-x?  The answer is simply the exponent, which removes e from the equation.

[atmyne]

... or whatever you said. You can always use www.wolframalpha.com if you're unsure of an answer... Just type the problem in. Like this:

wow what a great online resource! it even has some nifty black-scholes stuff that will come in handy for my derivatives class! nice one man.

[tjanuranus]

Stuck on something simple! How do you solve for.  Here.... (2/3)^x=81/16

[73109]

Log. I forget the formula.

[Aefenwelg]

81 = 3^4, 16 = 2^4

so, you have (2/3)^x = (3/2)^4

do the reciprocal, (2/3)^x = (2/3)^-4

x = -4

[tjanuranus]

Thank you I see what I was missing now! Also how do you work this problem out? I'm studying for a test tomorrow.

(1 + 0.10/12)^12t = 2

[Aefenwelg]

(1+.1/12)^12t = 2

take the ln of both sides:

12t*ln(1+.1/12) = ln2

divide:

t = ln2/(12*ln(1+.1/12))

[tjanuranus]

Ok you used the power formula first. I used the base changing formula and it worked but your way is probably better. Thank you again.

[tjanuranus]

Ok stuck on this one...

Log8x - log(1 + sqrt(x)) = 2.

Here is where I get to and I can't get the correct answer.

X = (100 + 100x^(1/2)) / 8


Either I'm doing it wrong up to that point or I'm wrong from the on.

[tjanuranus]

Ok I got the answer by putting ((8x) / (1+sqrt(x))) - 100 = y in the calculator and graphed it out. But algebraically still getting it wrong.

[Fiery Winds]

At this point:

X = (100 + 100x^(1/2)) / 8

Go a step back to put it into a form similar to a quadratic function.  From there, you can substitute to make it a quadratic function and solve using the quadratic formula.