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Solved - c2 logs

Hi everyone, i'm really confused on solving this question

It's from the C2 solomon paper H

The curves

(\frac{1}{3})^x

and

2(3^x)

intersect at the point P

So to find p I equated both of them and tried to use logs to solve the question

2(3^x)=(\frac{1}{3})^x

2x\log3 = x\log(\frac{1}{3})

But when i try to simplify this I end up going in circles and ending up with

xlog27=0

which isn't the answer i'm looking for.

How do you generally solve log equations with x's on both sides

HALP

(edited 12 years ago)

Reply 1

Dirac Spinor

note that:

(\frac{1}{3})^x=(3^{-1})^x=(3^x)^{-1}

now you have unknowns in terms of

3^x

. Can you solve that?

Reply 2

chemkid

Convert

(\frac{1}{3})^x

Unparseable latex formula:

3^-^x

Then separate it to form

Unparseable latex formula:

(\frac{3^x}{3^2^x})

Then say Let

y = 3^x

So on the LHS you get

(\frac{y}{y^2})

Then solve :smile:

(edited 12 years ago)

Reply 3

MinpoloD

Original post by ben-smith

note that:

(\frac{1}{3})^x=(3^{-1})^x=(3^x)^{-1}

now you have unknowns in terms of

3^x

. Can you solve that?

no because it still gives me xlog... on both sides

Reply 4

MinpoloD

I think i've got it another way, thanks for all your help guys

Reply 5

NewCrack

Unparseable latex formula:

(\frac{3^x}{3^2^x})

how did you get this???????

Reply 6

NewCrack

[QUOTE=Then separate it to form

Unparseable latex formula:

(\frac{3^x}{3^2^x})

i'm stuck on the same thing how did u get this??

Reply 7

olipal

Be careful of how you have taken logs.

2(3^x)=(\frac{1}{3})^x \Rightarrow log(2(3^x))=log(\frac{1}{3}^x)

\therefore log(2)+xlog(3)=-xlog(3)

Can you see what to do from here?

(edited 12 years ago)

Reply 8

Dirac Spinor

Original post by MinpoloD

no because it still gives me xlog... on both sides

If you have found your own way then great but you can do it the way I suggested and it really is the simplest method. I think you have misunderstood logs. rearrange to get

3^x=...

which you can do and then take the log of base 3 to find x, i.e.:

3^x=blah \Rightarrow log_3(3^x)=x=log_3(blah)

Reply 9

NewCrack

Original post by olipal

Be careful of how you have taken logs.

2(3^x)=(\frac{1}{3})^x \Rightarrow log(2(3^x))=log(\frac{1}{3}^x)

\therefore log(2)+xlog(3)=-xlog(3)

Can you see what to do from here?

i got to -Xlog3-Xlog3 = log2

then -Xlog (3/3) =log2

what do i do next???

Reply 10

Dirac Spinor

Original post by NewCrack

i got to -Xlog3-Xlog3 = log2

then -Xlog (3/3) =log2

what do i do next???

Careful with the negative signs

Reply 11

MinpoloD

Original post by ben-smith

If you have found your own way then great but you can do it the way I suggested and it really is the simplest method. I think you have misunderstood logs. rearrange to get

3^x=...

which you can do and then take the log of base 3 to find x, i.e.:

3^x=blah \Rightarrow log_3(3^x)=x=log_3(blah)

I solved it by simplifying and breaking down

(\frac{1}{3})^x= \dfrac{1^x}{3^x}

which enabled me to solve it by multiplying by

3^x

and have x on just one side rather than both.

I do understand logs, but in future i'll play around with the laws of indices before introducing logs to my equations