Help Please!

[Danielle058]

:? Okay, I'm having some trouble with this; if anyone can help it would be greatly appreciated! Thanks! (it's 3 different questions)

Solve for (x) to 3 decimal places.

1. ln(x+5)=ln(x-1)-ln(x+1)

2. 2.43 times 10^x=1.84

3. log base 3(-9)=x (in this problem I don't think it's possible because I've tried many times and can't find 3 to the ___ power that will give me a negative 9)

Thanks again for your Help!

 

[Matt]

Hi Danielle,

3. log base 3(-9)=x (in this problem I don't think it's possible because I've tried many times and can't find 3 to the ___ power that will give me a negative 9)

That's right, it's not possible. Namely because the argument of any logarithm must be positive. If there is a negative number in the argument (in this case -9), then the logarithm is undefined (in ther words, x is undefined since the argument of the logarithm, -9, is negative).

 

[ChaoticLlama]

1. ln(x+5)=ln(x-1)-ln(x+1)

Hint: ln(a/b) = ln(a) - ln(b)

2.
2.43 * 10^x = 1.84

10^x = 0.757201

x = log(0.757201)

x = -0.1207884

 

[soroban]

Hello, Danielle058!

Solve for (x) to 3 decimal places.

1. ln(x + 5) = ln(x - 1) - ln(x + 1)

We have: . ln(x + 5 ) + ln(x + 1) .= .ln(x - 1)

Now use the log property: . ln(a) + ln(b) .= .ln(ab)

[But I got two extraneous roots . . . the equation has <u>no</u> solutions.]

 

[Danielle058]

hm.. still need some help

:? Alright, still need a little help on this please, if anyone knows..
I get ln(x^2+6x+5)=ln(x-1) but how do you solve from there? Can you cancel out the In's and make it x^2+6x+5=x-1 and solve for x from there? Or.. what do u do in order to get the answers that you came too?

Thanks so much again.

Danielle.

 

[ChaoticLlama]

For the equation

ln(x^2+6x+5)=ln(x-1)

'Cancel' Is not the term to use. I hope you truely understand what to do, but if not here is a short explanation.

e^(ln(a)) = a

as does

ln(e^a) = a

So in effect, you are not 'cancelling' but you are raising the equation as a power of e. So you go like this

e^[ln(x^2+6x+5)] = e^[ln(x-1)]

x^2+6x+5 = x-1

TECHNICALLY you are 'cancelling' the lns, but that is a very unmathematical why to think about it.