solving a logarithmic equation

[hamilton77]

Hi,

I am stumped over a particular logarithmic equation. I keep getting a negative answer and even though it works apparently i cannot have a negative answer, so i'm confused.

the equation is

log(base 5)(3x) - log(base 5)(x - 1) = log(base 5)2

So I solved it like this:

log(base 5)(3x/x - 1) = log(base 5)2
(3x/x - 1) = 2
3x = 2(x - 1)
3x = 2x - 2
3x - 2x = -2
x = -2

When you plug -2 into it it works, doesn't it?


Thanks in advance

 

[mmm4444bot]

... apparently i cannot have a negative answer ...

... When you plug -2 into it it works, doesn't it?

Hello Hamilton:

Yes, the solution is x = -2.

Are you thinking that the solution cannot be negative because logarithms of negative numbers are not defined?

This is true in the Real Number System; however, we can take logarithms of negative numbers in the Complex Number System. Luckily for you, you do not need to know how to do that in order to find the solution to this equation because, as your method shows, you never needed to evaluate any of those logarithmic expressions.

I say "luckily" because I believe that logarithms in the Complex Number System involve infinite possibilities, so extra work is probably needed to properly evaluate them. (Don't quote me on this.)

Gobble, gobble,

~ Mark

 

[hamilton77]

Thanks for your reply. That's a plausible explanation, probably well beyond the scope of the current course i'm taking. So there's no conceivable way x could be evaluated differently? I'm not second-guessing you, just emphasizing that this course is only touching on log equations. None of the exercises we were assigned came out with a negative number.

Here's another:

log(base x + 6)14 = 4
(x + 6)^4 = 14
4th root of(x + 6)^4 = 4th root of 14
x + 6 = 1.93433642..
x = 1.93433 - 6
x = - 4.06566358..

 

[fasteddie65]

If you are in an Intermediate Algebra class, I would say that there is no solution to your first problem, because logarithms are defined in your course for positive numbers only. Correct?

 

[hamilton77]

If you are in an Intermediate Algebra class, I would say that there is no solution to your first problem, because logarithms are defined in your course for positive numbers only. Correct?

Yes, so I suppose I could write the answer and then 'not defined' in brackets or something to show I understand it can't be negative. Was there anything wrong with the second one i put up?

 

[hamilton77]

How about

2 log (x + 1) = 3
log (x + 1)^2 = 3
10^3 = (x + 1)^2
1000 = x^2 + 2x + 1
999 = x^2 + 2x
999 = x ( x + 2)
999/x + 2 = x

This one is obviously wrong. The book got 30.6 but that totally eludes me. Anyone?

 

[Denis]

2 log (x + 1) = 3
log (x + 1)^2 = 3

C'mon hamilton...
if 2 log (x + 1) = 3
then log(x + 1) = 3/2

 

[hamilton77]

2 log (x + 1) = 3
log (x + 1)^2 = 3

C'mon hamilton...
if 2 log (x + 1) = 3
then log(x + 1) = 3/2

Oh..that works, yup.

See, that's why I came here - to get help from people much smarter than myself :mrgreen:

Thanks