Logs

Original post by jacksonmeg

Idk :/ none of the rules I've got seem to work

Ignore log rules for now.
Just look at it as an algebra question :smile:

Reply 22

OP

Multiply out the brackets and solve x
but what would you get when you times log2 by x

(edited 10 years ago)

Reply 23

Original post by jacksonmeg

Multiply out the brackets and solve x

Exactly

Reply 24

OP

Idk how to expand it lol

Reply 25

Original post by jacksonmeg

Idk how to expand it lol

Use

\ a = \log_23

if it makes the expansion easier for you.
as TenOfThem says, it's just a number. :smile:

Reply 26

OP

Original post by joostan

Use

\ a = \log_23

if it makes the expansion easier for you.
as TenOfThem says, it's just a number. :smile:

I don't know what you mean, I don't really know anything about logs

Reply 27

OP

Is it like xlog2 + log2 -3xlog3 + 2log3?

Reply 28

TenOfThem

18

Original post by jacksonmeg

Is it like xlog2 + log2 -3xlog3 + 2log3?

yes

Reply 29

OP

I did. X -3x = - 1 -2
X= 1.5. But it's wrong

Reply 30

2710

18

Original post by jacksonmeg

I did. X -3x = - 1 -2
X= 1.5. But it's wrong

um...wut happened to your logs? Show your working

Reply 31

Original post by jacksonmeg

I did. X -3x = - 1 -2
X= 1.5. But it's wrong

Could you post everything you've done, cos it's hard to see where it is you're confused.

Reply 32

OP

Sry I'm typing on an iPad, I moved log2 and 2log3 to the other side making them a minus, then divded -log2by log2 leaving x on its own and -2log3 by log3 leaving -3x on its own

Reply 33

2710

18

Original post by jacksonmeg

Sry I'm typing on an iPad, I moved log2 and 2log3 to the other side making them a minus, then divded -log2by log2 leaving x on its own and -2log3 by log3 leaving -3x on its own

You cannot divide like that. You cannot divide one term and leave the other undivided lol.

You have to take x out as a factor and then move the factor across.

Reply 34

OP

I'll leave it then and ask in school tomorrow, need to go to sleep. Thanks for your help

Reply 35

Original post by jacksonmeg

Sry I'm typing on an iPad, I moved log2 and 2log3 to the other side making them a minus, then divded -log2by log2 leaving x on its own and -2log3 by log3 leaving -3x on its own

So you said:

\ (x+1)\log2 - (3x-2)\log3 = 0 [br]\Rightarrow x\log2 + \log2 - 3x\log 3 -2\log3 = 0

Now isolate x by factoring it out e.g

\ ax - bx = (a-b)x

Reply 36

OP

(log2-3log3)x + log2 -2log 3= 0 ?

Reply 37

OP

i did (log2-3log3)x = -log2 + 2log3
x = -log2 + 2 log3 / (log 2 - 3log3)

it didnt work :l what did i do wrong?

Reply 38

Original post by jacksonmeg

i did (log2-3log3)x = -log2 + 2log3
x = -log2 + 2 log3 / (log 2 - 3log3)

it didnt work :l what did i do wrong?

x =\dfrac{2\log3 - \log2}{\log2-3\log3}

? - It's very slightly wrong - a sign error has been made.
It's worth noting that using a different log base would give you an answer that looked different, but numerically would be identical. :smile:

Spoiler

(edited 10 years ago)

Reply 39

OP