logarithmic equations

i need some help on these equations. they may seem easy but i find it very difficult.

1. log 3/20 + log 5/12 - log 1/16

2. (2 - log 25)/(log square root of 8)

I dont know how to type the square root sign so i used words instead.

Hi there,

For the first question you need to use the fact that

log(a) + log(b) = log(ab) &
log(a) - log(b) = log(a/b)

So it simplifies to

log(16[3/20][5/12])
= log(240/240)
= log(1)


For the second question you probably need to make use of the fact that

log(x)^n = n.log(x)

but there are lots of ways of rearranging it. It is arguable which version is simpler - that depends on what you are trying to do.


All the best, Matt

OK thanks so far. For question 2 it asks us to simplify it. However this is the most I can simplify it, but I don't think it is right. Is there a way to simplify it further?

( 2 - 2 log 5 )/( 1/2 log 8 )

Well, if you factorise the numerator and multiply top and bottom by 2 you get

4 [1 - log5]/[log8]

4 {[1/log8] - [log5/log8]}

but 1/log8 = log8^(-1) = -log8

4 { -log8 - log5/log8 }

-4 {log8 + log5 - log8}

= -4 log5

Just a little question. Does log5/log8 equal to log5 - log8? I thought only log(5/8) equal to that. I'm probably wrong, after all log(5/8) may equal to log5/log8. Well anyway thanks a lot for your help.

Oh dear - yes you're quite correct

1/log(8) <> log(8^(-1))
1/log(8) = (log(8))^(-1)

log(8) = 3 * log(2)

ok so I suppose the most simplyfied equation will look like this then.
4 [ log 8^(-1) - (log5/3log2)]

I think this would be the simplest...
(4 / 3) * ((1 - log(5)) / log(2))

Well i dont know how you arrived at that answer so can you please explain.

2. (2 - log 25)/(log square root of 8)

= ( 2 - 2 log 5 )/( 1/2 log 8 )

= 4 [1 - log5]/[log8]

Which you already had. I'm assuming you are doing logs base 10 which makes

1 = log 10

so now have 4[log 10 - log 5]/[log 8]

= [4 log 2]/[3 log 2] using log(a) - log(b) = log(a/b) and log(8) = 3 log(2)

= 4/3

Usually if you are using logs and no base is quoted you are woking with base 10 (unless ln is written in which case you are using base e). Since this gives us a very nice answer I have no reason to doubt that we are using base 10 in this case. My working would give the same answer as kberry79 if I had not assumed we are working in base 10 and therefore left 1 alone.

1/log (8) does not equal log (8)^-1 so that method was incorrect

1/log (8) = 1 - log (8) = 1 + log (8)^-1

1 / log(8) can only be simplified to 1 / (3 * log(2)).

You can't do anything with exponents with 1 / log(8) except that
1 / log(8) = (log(8))^(-1)
(log(8))^(-1) is not the same as log(8^(-1)) so you cannot simplify it to -log(8)

1 / log(8) is not the same as 1 - log(8) either.

Good point with the base 10 assumed.

Assuming base 10, as mentioned by Glaysher above, it simplifies to 4 / 3.

1/ log 8 does not equal 1 - log 8 you're right. In my defence I was distracted by having my baby boy knaw off my hand (he's teething) as he is doing so right now

right, thank you all for helping. This certainly gives me a clearer understanding of logs.