[Perpetuallity]

Rewrite the equation in the form where c and d are constants.

I've tried numerous things but nothing seems to work. Some hints would be appreciated.

[nohomo]

Note that

[najinaji]

xlog3 = (x+1)log2

xlog3 = xlog2 + log2

xlog3 - xlog2 = log2

Am I on the right lines?

[nohomo]

I don't think you need logs to answer the question you've asked here OP

[Perpetuallity]

(Original post by nohomo)
I don't think you need logs to answer the question you've asked here OP

No, not really, but using natural logs I got it to

[shahruk]

(Original post by Perpetuallity)
Rewrite the equation in the form where c and d are constants.

I've tried numerous things but nothing seems to work. Some hints would be appreciated.

=> 3^x = 2 X 2^x

Dividing both sides by 2^x

=> (3^x)/(2^x) =2

=> (3/2)^x =2

=> 1.5^x = 2

Is that the answer ?

[steve2005]

(Original post by Perpetuallity)
No, not really, but using natural logs I got it to

I think your method is overkill.

[TheGrinningSkull]

(Original post by najinaji)
xlog3 = (x+1)log2

xlog3 = xlog2 + log2

xlog3 - xlog2 = log2

Am I on the right lines?

Then (Log rule)

[nohomo]

(Original post by Perpetuallity)
No, not really, but using natural logs I got it to

Steve2005 explains a simpler way of answering the question.

[steve2005]

(Original post by Perpetuallity)
No, not really, but using natural logs I got it to

Your 5.54 value should be 5,526...BUT no need to involve logs.

[Perpetuallity]

(Original post by steve2005)
Your 5.54 value should be 5,526...BUT no need to involve logs.

I know, I truncated (ln3-ln2) in my workings which carried through and changes the final answer slightly.

[Pride]

(Original post by steve2005)
I think your method is overkill.

I couldn't do it that way in an exam situation. I think I just don't know the indices rules to that level.

Oh wait, yes I do, normally we do it the other way round, multiply out the brackets, never tried that. nice...

[steve2005]

(Original post by Perpetuallity)
I know, I truncated (ln3-ln2) in my workings which carried through and changes the final answer slightly.

I realise that is what you did BUT it is not correct to round as you progress through a calculation.

ROUNDING should only occur at the final answer stage.

[najinaji]

(Original post by TheGrinningSkull)
Then (Log rule)

Excellent! Hoped that would be the right method.

[Perpetuallity]

(Original post by steve2005)
I realise that is what you did BUT it is not correct to round as you progress through a calculation.

ROUNDING should only occur at the final answer stage.

I'll use the "ans" button on the calculator next time. Thanks for the help.

[TheGrinningSkull]

(Original post by najinaji)
Excellent! Hoped that would be the right method.

Yea, it looks right, it makes sense as well, and then you can solve for x