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

1. ## 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

2. (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.

3. 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.

4. (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?

5. 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.

6. 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.

7. thanks

8. 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?

9. 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.

10. Sorry. I didn't check back to this site since I posted. I wound up getting the problems. Thanks for the help.