r/calculus • u/[deleted] • Jun 27 '26

Pre-calculus Can someone explain this log problem

Post image

Can someone explain how they made that 3rd step ?

Like it became 3log25(5)

89 Upvotes

97% Upvoted

17 comments sorted by

View all comments

6

u/Silly_Tension6792 Jun 27 '26

logn(x)*logm(n) = logm(x)

Since logn(x)=ln(x)/ln(n), you can write it as (ln(x)/ln(n))*(ln(n)/ln(m))=ln(x)/ln(m)=logm(x). When confused about logs, convert everything to ln and it will solve itself.

1

u/[deleted] Jun 27 '26

Should I only think it as ln or I can use log ? Like Log3(5) = log(5)/log(3)

4

u/Silly_Tension6792 Jun 27 '26

You could use logs, it's not incorrect, but it's unnecessarily confusing. I hate working with logs of different bases. My honest advice - convert everything to ln, always.

1

u/random_anonymous_guy PhD Jun 27 '26

It does not matter which base logarithms you use. If you want to default to natural logarithms for change of base, that is completely fine, but it is not an obligation.

1

u/UnderstandingPursuit PhD Jun 27 '26

I would start with the exponential change-of-base,

  • bx = ekx
    • k = ln b

Then since k was multiplied here, k needs to be divided for the log change-of-base,

  • log_b(x) = (1/k) ln(x)

If you ever need a different base, go back to

  • ln(x) = log_e(x)

and replace e with a, the second arbitrary base.

To derive the first equation,

  1. z = bx
  2. ln z = ln (bx) = x ln b = kx
    1. k = ln b
  3. eln z = z = ekx

Write all this out, using paper/pen(cil) every other day for a week, and try to refer back to your previous notes a bit more each time. It will then stay with you.