I am stuck on this prblem (logarithms)

[abel muroi]

I was given the problem log₅(x + 1) - log₅(x - 1) = 2 and was told to solve for x

here is the work i have so far...

log₅(x + 1) - log₅(x - 1) = 2
log₅((x + 1)/(x - 1)) = 2

Im not sure what to do next here, should i take the log of 2 as well? or should i divide log₅ with 2?

 

[wjm11]

I was given the problem log₅(x + 1) - log₅(x - 1) = 2 and was told to solve for x

here is the work i have so far...

log₅(x + 1) - log₅(x - 1) = 2
log₅((x + 1)/(x - 1)) = 2

Im not sure what to do next here, should i take the log of 2 as well? or should i divide log₅ with 2?

Try converting from logarithmic form into exponential form.
Example: http://virtualnerd.com/algebra-2/ex...ithms-introduction/convert-log-to-exponential

Hope that helps.

 

[daon2]

I was given the problem log₅(x + 1) - log₅(x - 1) = 2 and was told to solve for x

here is the work i have so far...

log₅(x + 1) - log₅(x - 1) = 2
log₅((x + 1)/(x - 1)) = 2

Im not sure what to do next here, should i take the log of 2 as well? or should i divide log₅ with 2?

Hint: 2 = log₅25

 

[abel muroi]

Hint: 2 = log₅25

Thank you! that hint was all i needed!

 

[abel muroi]

Hint: 2 = log₅25

What about the problem log x + log (x - 3) = 1

should i use log for 1?

 

[daon2]

What about the problem log x + log (x - 3) = 1

should i use log for 1?

1 = log_N(N) for any valid base N.

 

[abel muroi]

1 = log_N(N) for any valid base N.

wait i think i have it figured out, can you check if i did this right?

log x + log (x - 3) = 1
log (x² - 3x) = log 0
x² - 3x = 0
(x - 3) (x + 0) = 0

x = 3

did i do this correctly?

 

[Ishuda]

wait i think i have it figured out, can you check if i did this right?

log x + log (x - 3) = 1
log (x² - 3x) = log 0
x² - 3x = 0
(x - 3) (x + 0) = 0

x = 3

did i do this correctly?

No, you went backwards: log(1)=0, not log(0)=1!

If the log is the natural log [base e], then
log (x² - 3x) = log e
x² - 3x = e

 

[abel muroi]

No, you went backwards: log(1)=0, not log(0)=1!

If the log is the natural log [base e], then
log (x² - 3x) = log e
x² - 3x = e

So once i get the e, how can i factor x² - 3x - e = 0?

 

[abel muroi]

how can i factor x^2 - 3x - e = 0?

Im not used to factoring an e in a equation, can someone show me the way of a mathematician?

 

[Deleted member 4993]

Im not used to factoring an e in a equation, can someone show me the way of a mathematician?

Treat e as a constant (e ≈ 2.718) in a quadratic equation.

 

[Mrspi]

What about the problem log x + log (x - 3) = 1

should i use log for 1?

In the "olden days" when I was studying math in college, when no base was specified for "log", the log was assumed to be a "common logarithm" with a base of 10.

So, log 100 = 2, since 100 = 10²

So.....I would handle your problem this way:

log x + log (x - 3) = 1

log (x^2[ - 3x) = log 10¹

x² - 3x = 10

x² - 3x - 10 = 0

Now....that factors nicely! Ball is in your court.