# Math Help - Natural Exponential Function

1. ## Natural Exponential Function

Find the equation of the tangent to the curve defined by $
y=x-e^{-x}$
that is parallel to the line represented by 3x-y-9=0

This is my work, can someone check it please? I'm not sure if i have the rules of the natural exponential function correct.

First off I want to say, I know that $e^{lnm} =m$
BUT

is there anything for $e^{1/lnm}=??$

Now I will continue.
$y=x-e^{-x}$
$y'=1+e^{-x}$

y'=3 at x=?

$3=1+e^{-x}$
$ln2=lne^{-x}$
$ln2=-xlne$
x=-lne/ln2
x=-1/ln2

y=mx+b
$y=(-1/ln2)-e^{-(-1/ln2)}$
$y=(-1/ln2)-e^{(1/ln2)}$

y=mx+b

3(-1/ln2)+b= $(-1/ln2)-e^{(1/ln2)}$
b= $(2/ln2)-e^{(1/ln2)}$

y=3x+ $(2/ln2)-e^{(1/ln2)}$

2. Originally Posted by skeske1234
Find the equation of the tangent to the curve defined by $
y=x-e^{-x}$
that is parallel to the line represented by 3x-y-9=0
$y = x - e^{-x}
$

$y' = 1 + e^{-x}$

$1 + e^{-x} = 3$

$e^{-x} = 2$

$-x = \ln(2)$

$x = -\ln(2)$

note that $-\ln(2) \ne \frac{1}{\ln(2)}$ ,
$-\ln(2) = \ln(2)^{-1} = \ln\left(\frac{1}{2}\right)$

to get the equation of the tangent line ...

at $x = -\ln(2)$

$y = -\ln(2) - e^{\ln(2)} = -[\ln(2) + 2]$

$y + [\ln(2) + 2] = 3[x + \ln(2)]$

$y = 3x + 2[\ln(2) - 1]$

3. It might be easier if you use y = m(x-x0) +y0 instead of y = mx + b

no you can't simplify e^(1/ln(m))

4. Originally Posted by skeeter
$y = x - e^{-x}
$

$y' = 1 + e^{-x}$

$1 + e^{-x} = 3$

$e^{-x} = 2$

$-x = \ln(2)$

$x = -\ln(2)$

note that $-\ln(2) \ne \frac{1}{\ln(2)}$ ,
$-\ln(2) = \ln(2)^{-1} = \ln\left(\frac{1}{2}\right)$

to get the equation of the tangent line ...

at $x = -\ln(2)$

$y = -\ln(2) - e^{\ln(2)} = -[\ln(2) + 2]$

$y + [\ln(2) + 2] = 3[x + \ln(2)]$

$y = 3x + 2[\ln(2) - 1]$
I'm not too sure of your second last line.. is there another way of putting it?

5. Originally Posted by skeske1234
I'm not too sure of your second last line.. is there another way of putting it?
$y - y_1 = m(x - x_1)$

6. Originally Posted by skeeter
$y - y_1 = m(x - x_1)$
ok.. why do we have to use that here? if I was to use y=mx+b (my teacher uses this in class) how would it look? I am trying to figure it out using that way
my work:

y=mx+b
-(ln2+2)=3(-ln2)+b
-ln2-2=-3ln2+b
-ln2+3ln2-2=b
-ln2+ln8-2=b
ln6-2=b

edit: if this is still wrong, can you please show me using the y=mx+b formula? thanks

7. sorry, still seem to be having some trouble with this one>>
my result ends up to be....

y=3x+ln6-2

is this still wrong?

8. Originally Posted by skeske1234
ok.. why do we have to use that here? if I was to use y=mx+b (my teacher uses this in class) how would it look? I am trying to figure it out using that way
my work:

y=mx+b
-(ln2+2)=3(-ln2)+b
-ln2-2=-3ln2+b
-ln2+3ln2-2=b
-ln2+ln8-2=b
ln6-2=b

edit: if this is still wrong, can you please show me using the y=mx+b formula? thanks
you need to review your algebra for logs ...

$\ln(8) - \ln(2) \ne \ln(6)$

$\ln(8) - \ln(2) = \ln\left(\frac{8}{2}\right) = \ln(4)$