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C2 Logarithms question- driving me crazzzy! Please help!

13

OK so this is the question that I'm stuck on:

Given that a = lg2 and b = lg3, express each of the following in terms of a and b:

h) lg60 + lg20 - 2

Btw, this has to be done without using a calculator, just using the laws of logarithms, and lg means log base 10.

I can put up some of my working for other parts of the question if that would help or if you'd like to check them for me? :p:

:p:

:redface:

But it's the question above that's really getting under my skin and I've been trying to do for some while now. :redface:

:redface:

Elements xXx

Reply 1

5

\lg60 + \lg20 - 2 = \lg1200 - \lg100 = \lg12

However, I suspect the question asked for

a=\lg2, b=\lg9

or

b=\lg3

Because

3\lg2=\lg8, \text{ so } 3a=b

The question wont ask you to express lg60 + lg20 - 2 in terms of a and b since it can be expressed solely in a or b.

Reply 2

10

I don't think that

b = \mathrm{log}\;8

because as khaixiang said, it can be expressed in the form of

a

.

However, for the first part, where

Unparseable latex formula:

\mathrm{log} = \mathrm{log}\,_{10}

:

\mathrm{log}\;20 = \mathrm{log}\;10 + \mathrm{log}\;2.

Therefore,

\mathrm{log}\;20 = 1 + \mathrm{log}\;2.

I'd imagine that you do the same for

\mathrm{log}\;60

, provided the value of

b

was different. :smile:

:smile:

Reply 3

Perhaps its a misprint in the question? Becase log8 = log(2^3) = 3log2 so it wouldnt make sense since for b = log8, then b = 3a

Reply 4

OP

13

lg12 = lg2 + lg2 + lg3
lg12 = 2a + b

Reply 5

OP

13

Oh yes, I'm ever so sorry! I wrote the question out correctly on my maths book but wrote it out incorrectly on here! :redface:

:redface:

:redface:

b is actually lg3.

Reply 6

OP

13

First post edited. :redface:

:redface:

I feel so embarassed now after that typo. :redface:

:redface:

:redface:

:vroam:

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