ln and ex log questions

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December 17th 2006, 10:37 AM
SportfreundeKeaneKent

ln and ex log questions

Here are some ln (lawn) and e^x questions. I'm still learning these so I'm not so good at solving thema and these ones are hard so I was wondering if anyone can help.

http://img329.imageshack.us/img329/3...p141922wy7.png

If that picture doesn't work then here's a link
http://img329.imageshack.us/img329/3...p141922wy7.png
December 18th 2006, 07:04 AM
Hootenanny

How much do you know about logarithms? To me these questions appear to be elementary applications of the laws of logs. Could you be more specific as to where you're having problems?
December 18th 2006, 08:02 AM
CaptainBlack

Solve $e^x-3e^{-x}=2$

Multiply through by $e^x$ and rearrange to get:

$e^{2x}-2e^x-3=0$

Now put $u=e^x$ and you get a quadratic:

$u^2-2u-3=0$ .

Solve this and then the solutions for $x$ are the natural logarithms of the $u$ 's.

The seventh question - Solve $e^x+7e^{-x}=8$
can be solved by the same method.

RonL
December 18th 2006, 08:58 AM
Soroban

Hello, SportfreundeKeaneKent!

Here are a couple of them . . .

Quote:

2) Solve: . $(\ln x)^2\:=\:\ln(x^2)$

We have: . $(\ln x)^2\:=\:2\ln x\quad\Rightarrow\quad (\ln x)^2 - 2\ln x \:=\:0$

Factor: . $\ln x(\ln x - 2)\:=\:0$

And we have two equations to solve:

. . $\ln x \,= \,0\quad\Rightarrow\quad\boxed{ x \,= \,1}$

. . $\ln x \,=\,2\quad\Rightarrow\quad\boxed{ x \,= \,e^2}$

Quote:

3) The graph of $y = \ln x$ is rotated 90° CCW about the origin.
What is the equation of the new graph?

The graph of $y = \ln x$ looks like this:

Code:

| | * | * ------+-------*---------- | * 1 | * | * | |* |

Rotated 90° CCW, it looks like this:

Code:

* | | * | * | * | * | * | * | * ------------+-------------------- |

We're expected to recognize this as the graph of: $y \:=\:e^{-x}$
December 18th 2006, 09:12 AM
SportfreundeKeaneKent

We're sort of doing the log unit at school right now. I thought that these seemed pretty simple too but there's always one extra thing you have to do on some of these harder problems that the teacher gives. The ones from the textbook aren't a problem. lnx is a bit more of a problem and I'm still learning that.
December 18th 2006, 09:44 AM
earboth

1 Attachment(s)

Quote:

Originally Posted by SportfreundeKeaneKent

We're sort of doing the log unit at school right now. I thought that these seemed pretty simple too but there's always one extra thing you have to do on some of these harder problems that the teacher gives. The ones from the textbook aren't a problem. lnx is a bit more of a problem and I'm still learning that.

Hello Sportsfreund,

I'll send you a .pdf-file to show you how the rules of powers and the rules of logarithm are connected. (It isn't necesary to learn German to understand this file)

EB
December 18th 2006, 10:00 AM
Soroban

Hello again, SportfreundeKeaneKent!

A couple more . . .

Quote:

4) Solve: . $\ln\left(\frac{e^2x^2+1}{2x}\right) \:=\:1$

We have: . $\frac{e^2x^2+1}{2x} \:=\:e\quad\Rightarrow\quad e^2x^2 + 1 \:=\:2ex\quad\Rightarrow\quad e^2x^2 - 2ex + 1 \:=\:0$

Factor: . $(ex - 1)^2\:=\:0\quad\Rightarrow\quad ex - 1\:=\:0\quad\Rightarrow\quad ex \,=\,1\quad\Rightarrow\quad x \,=\,\frac{1}{e}$

Quote:

6) Solve: . $\ln(x^2 + e) - \ln(x + 1)\:=\:1$

We have: . $\ln\left(\frac{x^2+e}{x+1}\right) \:=\:1\quad\Rightarrow\quad\frac{x^2+e}{x+1} \:=\:e\quad\Rightarrow\quad x^2 + e \:=\:ex + e\quad\Rightarrow\quad x^2 - ex \:=\:0$

Factor: . $x(x - e)\:=\:0$

And we have two equations to solve:

. . $\boxed{x\:=\:0}$

. . $x - e \:=\:0\quad\Rightarrow\quad\boxed{x = e}$

and both answers check out . . .
December 19th 2006, 06:06 PM
SportfreundeKeaneKent

Could someone check up on these answers for me for these questions:

For question 1, I got x = 1.098

For question 5, I got x = 1.946 and x = 0

For question 2, does x = 0 AND e^x or just e^x/7.389

And for the last question (e^3x=15), I got x = 0.9026
December 20th 2006, 01:48 PM
SportfreundeKeaneKent

Oh and also, I was doing question 4 down that list and I got x = 0.5 which is different than the answer Soroban's answer of 1/e. I was wondering if anyone could check up on this.

I used the quotient rule in this question so that it became this:
ln(e^2 x^2+1)-ln2x=1
2lne^2 x+ 0 - ln2x=1
2x + 0 - ln2x = 1
2x(1-ln1)=1
2x=1
x-1/2 or 0.5

When I plugged x = 0.5 into the equation to check my work, it worked out so I'm really confused about wheter what Soroban did was right or the way I've done it is right.
December 20th 2006, 11:31 PM
earboth

Quote:

Originally Posted by SportfreundeKeaneKent

Oh and also, I was doing question 4 down that list and I got x = 0.5 which is different than the answer Soroban's answer of 1/e. I was wondering if anyone could check up on this.

I used the quotient rule in this question so that it became this:
ln(e^2 x^2+1)-ln2x=1
2lne^2 x+ 0 - ln2x=1
2x + 0 - ln2x = 1
2x(1-ln1)=1
2x=1
x-1/2 or 0.5

When I plugged x = 0.5 into the equation to check my work, it worked out so I'm really confused about wheter what Soroban did was right or the way I've done it is right.

Hello Sportsfreund,

here:

ln(e^2 x^2+1)-ln2x=1
2lne^2 x+ 0 - ln2x=1

you've made a very common mistake: You used the "ln" as a kind of variable and expanded the brackets by rule of distribution. But ln is a function and the contents of the brackets are the arguments of this function.
You can get rid of the "ln" if you use the following property:

$e^{\ln(x)}=x$

Soroban's solution is 100% correct.

EB