Author Topic: solving for x. (Read 1122 times) Tweet Share

nbalakers24 · « **on:** March 29, 2010, 07:56:48 pm »

hiii

i need help solving this by hand:

1-loge(x+1)

Cheers!

m@tty · « **Reply #1 on:** March 29, 2010, 08:00:13 pm »

I'll assume it is equal to, say, y.

$y=1-\log_e{(x+1)}$

$-(y-1)=\log_e{(x+1)}$

$e^{-(y-1)}=x+1$

$x=\frac{1}{e^{y-1}}-1$

and if y=0 then

$x=\frac{1}{e^{-1}}-1$

$x=e-1$

EDIT: leet

nbalakers24 · « **Reply #2 on:** March 29, 2010, 08:02:38 pm »

cheers mateeee !

nbalakers24 · « **Reply #3 on:** March 29, 2010, 08:03:18 pm »

haha you had 222 karma, and i have 22 karma

m@tty · « **Reply #4 on:** March 29, 2010, 09:35:47 pm »

Oh yeah, something you should ALWAYS check is the validity of solutions, by substituting them in the original equation. Another thing to watch out for is if the calculated solutions are in the domain of the original equation.

So in this case $y=1-\log_e{(x+1)}$

For the logarithm $x+1>0$

$x>-1$

$x\in(-1,\infty)$

For the solution, $x=e-1$ , to be valid it must fall within this domain. And it does. Next you should check the answer by direct substitution.

$sub \ x=e-1,$

$1-\log_e{(e-1+1)}$

$1-1=0$

Therefore the point $(e-1,0)$ is a point on the curve $y=1-\log_e{(x+1})$

the.watchman · « **Reply #5 on:** March 30, 2010, 08:54:46 am »

Yep, good job m@tty
Checking solutions is really important for logarithmic questions, or really anything with an implied domain that isn't R.
It's a very easy way to lose marks

Mao · « **Reply #6 on:** March 30, 2010, 11:51:55 am »

I don't see a particular reason to transpose $y = 1 - \log_e (x+1)$ to in terms of y, you can solve it directly:

$\begin{align*} 0 & = 1 - \log_e(x+1) \\ \log_e (x+1) & = 1 \\ x + 1 & = e\\ x & = e - 1\\ & \approx 1.718..\\ \end{align*}$

m@tty · « **Reply #7 on:** March 30, 2010, 12:01:42 pm »

Yes, but he said solve $1-\log_e{(x+1)}$ . Since he did not specify what this is to be equal to I gave a general solution where it is equal to the generic variable y.

nbalakers24 · « **Reply #8 on:** April 04, 2010, 01:42:35 pm »

ty guys. got another question.

find all values of x between 0 and 2pie

sinx=-0.45

i need help :S

thanks guys

nbalakers24 · « **Reply #9 on:** April 04, 2010, 02:02:32 pm »

dont worry ^^^^

was not in right mode. lol.

Stroodle · « **Reply #10 on:** April 04, 2010, 03:06:08 pm »

lol. a guy did that in our spesh exam the other day, and half of it was trig. can't believe he didn't realise...

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