# Logarithms..

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• May 8th 2007, 06:10 AM
Ruler of Hell
Logarithms..
log(10 + 10^1/3)

^ = Raised to (Exponentiation)

Another few should be coming your way. Please, I need help urgently!
• May 8th 2007, 06:12 AM
topsquark
Quote:

Originally Posted by Ruler of Hell
log(10 + 10^1/3)

^ = Raised to (Exponentiation)

Another few should be coming your way. Please, I need help urgently!

Unless you are going to approximate it in some way, there's nothing you can do with it.
log(a + b)
has no expanded form in terms of a and b.

(Note that log(ab) = log(a) + log(b).)

-Dan
• May 8th 2007, 06:18 AM
janvdl
Well its log(10^1 + 10^1/3)
Cant we get the 1 and the 1/3 out of there somehow? :confused: However, it doesnt really seem likely to me.
• May 8th 2007, 06:20 AM
Jhevon
Quote:

Originally Posted by Ruler of Hell
log(10 + 10^1/3)

^ = Raised to (Exponentiation)

Another few should be coming your way. Please, I need help urgently!

i was going to make the same comment as topsquark. there is no nice algebraic way to break this down, so unless you can use log tables or a calculator (which i don't think is the idea), it's near impossible to do this. are you sure it shouldn't be 10*10^(1/3)
• May 8th 2007, 06:22 AM
janvdl
Quote:

Originally Posted by Jhevon
i was going to make the same comment as topsquark. there is no nice algebraic way to break this down, so unless you can use log tables or a calculator (which i don't think is the idea), it's near impossible to do this. are you sure it shouldn't be 10*10^(1/3)

Yes, are you sure you didnt make a typo?

If it should be 10*10^(1/3) then
log(10 x 10^1/3) = log(10) + log(10^1/3)
= 1 + 1/3
= 4/3
• May 8th 2007, 06:24 AM
Ruler of Hell
That's what I was thinking. It has to be a mistake with the sum, I suppose.

PROVE
log4 + log2 = log(to the base 3)9
• May 8th 2007, 06:28 AM
janvdl
Quote:

Originally Posted by Ruler of Hell
That's what I was thinking. It has to be a mistake with the sum, I suppose.

PROVE
log4 + log2 = log(to the base 3)9

RHS: log_3 9 = 2

LHS: log_10 4 + log_10 2 = log_10 8 = 0,9
• May 8th 2007, 06:33 AM
Ruler of Hell
I didn't understand you. I got RHS as 2, and I got LHS as 3log2. Umm... How is that proved?
• May 8th 2007, 06:34 AM
Jhevon
Quote:

Originally Posted by Ruler of Hell
That's what I was thinking. It has to be a mistake with the sum, I suppose.

PROVE
log4 + log2 = log(to the base 3)9

as you can probably tell from janvdl's answer, this is not true. maybe the book made another typo:D
• May 8th 2007, 06:37 AM
janvdl
Quote:

Originally Posted by Ruler of Hell
I didn't understand you. I got RHS as 2, and I got LHS as 3log2. Umm... How is that proved?

3log_10 2 = log_10 8 :)
• May 8th 2007, 06:38 AM
Ruler of Hell
Not a typo, incorrect sums given in the book. Thanks for the help!
• May 8th 2007, 06:40 AM
Jhevon
Quote:

Originally Posted by Ruler of Hell
Not a typo, incorrect sums given in the book. Thanks for the help!

i'm seriously considering telling you to through that book away:mad: but if that's what your school uses, i guess you can't do that:D

books make mistakes once in a while, but usually not often enough for you to pin point two in the same section, and such blatant mistakes to top it all off
• May 8th 2007, 06:42 AM
topsquark
Quote:

Originally Posted by Ruler of Hell
That's what I was thinking. It has to be a mistake with the sum, I suppose.

PROVE
log4 + log2 = log(to the base 3)9

Please start a new thread for a new problem.

If no base is given log(4) can be either of two things, depending on who is teaching the class and what field you are in. It could either be
log_{10}(4)
or
log_e(4) = ln(e)

However as log_3(9) = log_3(3^2) = 2 I can think of only one way such that this is going to work:

log_b(4) + log_b(2) = 2

log_b(4*2) = 2

log_b(8) = 2

Using the change of base formula:
log_b(8) = ln(8)/ln(b) = 2

ln(b) = (1/2)ln(8)

b = e^{(1/2)ln(8)} = [e^{ln(8)}]^{1/2} = sqrt{8} = 2.82843

So we CAN say:
log_{sqrt{8}}(4) + log_{sqrt{8}}(2) = log_3(9)

but I doubt this is what the question was after.

-Dan