Logarithms (1 Viewer)

[sasquatch]

In regards to logarithms, are we allowed to use the identities:

log_a(xy) = log_ax + log_ay
log_a(x/y) = log_ax - log_ay
log_a(xⁿ) = n log_ax

Or do we have to establish them before using... Just asking as my book does "box" them as it does with many other rules, formulae, identities, ect.

 

[Riviet]

In regards to logarithms, are we allowed to use the identities:

log_a(xy) = log_ax + log_ay
log_a(x/y) = log_ax - log_ay
log_a(xⁿ) = n log_ax

Or do we have to establish them before using... Just asking as my book does "box" them as it does with many other rules, formulae, identities, ect.

You don't need to establish these identities or any of the log laws at all, but textbooks like to do this so you know what it's illustrating in a worked example. It might be in a line of working such as: " =(log2x3), by log law 1 ", just so you can see how they got to that line.

So when you're doing questions that involve any log laws, you don't need to state anything, of course you still need to show the normal working in your answer.

 

[SoulSearcher]

In regards to logarithms, are we allowed to use the identities:

log_a(xy) = log_ax + log_ay
log_a(x/y) = log_ax - log_ay
log_a(xⁿ) = n log_ax

Or do we have to establish them before using... Just asking as my book does "box" them as it does with many other rules, formulae, identities, ect.

Absolutely not.

You do not need to prove any of log laws before you can use it. They just show the proof so that you can understand where it came from

 

[sasquatch]

Wait so if a question asks..erm..

Given log_e10 = 2.3026 and log_e7 = 1.9459 Find log_e0.007

Can you answer it just as:

log_e0.007
= log_e(7 / 1000)
= log_e7 - log_e1000
= log_e7 - log_e10³
= log_e7 - 3log_e10
= whatever (cant be botehred)

or do you have to do it as...

log_e0.007
= log_e(7 / 1000)

Let e^x = 7, x = log_e7
Let e^y =1000, y = log_e1000

e^x / e^y = e^{x - y}

.:. log_e(e^x/e^y) = x - y
= log_e7 - log_e1000
= log_e7 - log_e10³

Let 10³ = e^z, z = log_e10³

10 = e^z/3

z/3 = log_e10
z = 3log_e10

.:. log_e10³ = 3log_e10

hence log_e0.007 = log_e7 - 3log_e10
= some decimal

 

[Riviet]

Hell no! Just do it the first way you had, you don't have time to be doing this in an exam or assessment and it isn't necessarily since the log rules form the basics of logarithms.

 

[sasquatch]

thank god.. it took me so long to write that up.. its hard doing maths through typing, especially when you cant see what your last steps were..gets confusing with all those [su p], [su b] ect. tags..

 

[SoulSearcher]

unless they specifically ask for the proof, then you don't have to do it. It just wastes valuable exam time.

 

[sasquatch]

cool then..