# Few HW problems involving Logs, Natural logs, and Continuity.

• February 12th 2011, 03:38 PM
woody189
Few HW problems involving Logs, Natural logs, and Continuity.
Hi. I'm having problems with a few HW problems.

1) log5 (x) + log5 (x+4) = ?

How would I solve this? I get as far as log5 x^2+4x, but then I'm lost.

2) (4x^3 - 5x^2 + 8x -7)/ ( x-1) for x<1
-3x^2 +3x + a for x>=1

solve for a.

I can't seem to get rid of the (x-1) successfully

3) e^(x+3) = e^(x) +7

solve for x

This one seems like it should be simple , but I can't solve it past

x^2 +3x = ln7

I don't really need the answers to these. I would just like to know how to solve them. I've been stuck on them for hours.

Thanks
• February 12th 2011, 04:08 PM
General
(1)
there is no equation!
what is the right hand side?

(2)
Solve what?
I do not see an equation. There is only a function!

(3)
Taking 'ln' of both sides will not solve it.
• February 12th 2011, 04:26 PM
woody189
Quote:

Originally Posted by General
(1)
there is no equation!
what is the right hand side?

(2)
Solve what?
I do not see an equation. There is only a function!

(3)
Taking 'ln' of both sides will not solve it.

Wow.. I completely screwed that up.... Sorry about that.

Let's try that again:

1) log5 (x) + log5 (x+4) = k
solve for x in terms of k.

2) (4x^3 - 5x^2 + 8x -7)/ ( x-1) for x<1
-3x^2 +3x + a for x>=1

What value for a will make this function continuous at 1?

3) e^(x+3) = e^(x) +7

For this I first brought the e's to one side so I got:

e^(x+3) - e^(x) = 7

Then I combined them using log(x) - log(y) = log(x/y). I think that's allowed in this case, is it not? After that I got:

e^((x+3)/x) = 7

Then I just used ln to get:
(x+3)/x = ln7

I must have done something wrong, I just don't know what.
• February 12th 2011, 04:41 PM
General
(1)
You did the hard job :
$\log_5 (x^2+4x) = k$

We want to remove this noisy log. To do that, you need to apply this formula: $a^{log_a{x}}=x$. Got it?

(2)
Let f(x) be our function. f is continuous at x=1 $\displaystyle \implies f(1) = \lim_{x \to 1^-} f(x)$

(3)
Oh my god!
No No No No No. $\displaystyle e^a - e^b \neq e^{\frac{a}{b}}$

The property $log(a)-log(b)=log(\frac{a}{b})$ is valid for the logarithmic functions only, not for the exponential functions.

To solve your eqaution, re-write it as : $e^3 \cdot e^x - e^x = 7$ . Can you complete it now?
• February 12th 2011, 05:33 PM
woody189
Okay. So for problem 1, I got that far, but it just looked funny and I couldn't figure out what to do. The 5^k is what really threw me off. I was trying to get a an integer answer I guess, and gave up. After you let me know I was on the right track, I got the right answer.

2) I got that, and I get the 0/0 which of course means I need to do some work to get rid of the 0/0. However, I can't seem to figure out what to do. I can't get rid of the (x-1). I'm gonna keep working on it though.

3) Lol... sorry for startling you. Now that I look at it, it makes a lot more sense to do what you did. However, sometimes it's just hard for me to see stuff, even though I know the rules, I just can't see that that rule should apply in this situation. At an rate, I got the answer for that one as well.

Thanks for the help. Hopefully I get number 2 on my own, if not, I'll be back.

Also, Is there a sticky for how to post like you did? like get the square root sign instead of typing sqrt, or showing an exponent instead of the carrot (^)? I'm sure I'll be back on these forums, and I think it's easier for everyone to view it like that.
• February 12th 2011, 08:12 PM
mr fantastic
Quote:

Originally Posted by woody189
[snip]
Also, Is there a sticky for how to post like you did? like get the square root sign instead of typing sqrt, or showing an exponent instead of the carrot (^)? I'm sure I'll be back on these forums, and I think it's easier for everyone to view it like that.

Click on the link in my signature.
• February 12th 2011, 11:43 PM
woody189
thanks
• February 13th 2011, 05:01 AM
General
For (1) : You will not get an integer answer. You will get an answer in terms of k.

For (2) : Did you know how to deal with the 0/0 one?
• February 13th 2011, 08:01 AM
rtblue
for number one:

$\displaystyle \log_5 (x^2+4x) = k$

$\displaystyle 5^k=x^2+4x$

$\displaystyle x^2+4x-5^k=0$

Try using the quadratic formula from here.
• February 22nd 2011, 05:39 PM
woody189
Sorry. I didn't check back to this site since I posted. I wound up getting the problems. Thanks for the help.