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Year 12s: what will be the first way you research your future options? >>

Maths C2 help

Hello I am really struggling with this question, wonder could anyone solve this for me?
log6 + log(x-3) =2 log y
2y -x =3

thanks

===Moved to Maths===

(edited 7 years ago)

Original post by Marcus2016

Hello I am really struggling with this question, wonder could anyone solve this for me?
log6 + log(x-3) =2 log y
2y -x =3

thanks

===Moved to Maths===

I'm not going to solve it for you, but I will give you a starting step/hint:

Can you use the equation

\displaystyle \log 6 + \log \left(x-3\right) = 2 \log y

to express

y

in terms of

x

, or just an equation linking the two?

Then it's just solving simultaneous equations from C1!!

OP

Original post by crashMATHS

I'm not going to solve it for you, but I will give you a starting step/hint:

Can you use the equation

\displaystyle \log 6 + \log \left(x-3\right) = 2 \log y

to express

y

in terms of

x

, or just an equation linking the two?

Then it's just solving simultaneous equations from C1!!

Thanks but when i multiply log into the bracket what does it become, is it log -3, as you can't get a negative log?

remember....

log p + log q = log pq

alog p = log {p^a}

Original post by Marcus2016

Thanks but when i multiply log into the bracket what does it become, is it log -3, as you can't get a negative log?

Oh no, you have a misunderstanding. It is generally not true that

\log (a+b) = \log(a) + \log(b)

. That is not how logarithms work.

What you want to be using is that:

\log(a) + \log(b) = \log(ab)

for the left hand side. In this case, let your

a = 6 , b = x-3

OP

Thanks everyone question solved :smile:

:smile:

I owe you one!

OP

sorry to ask again but any hints for the following?

2 + log2(2x +1) =2 log2 y
x=22-y

Study Forum Helper

Original post by Marcus2016

sorry to ask again but any hints for the following?

2 + log2(2x +1) =2 log2 y
x=22-y

2=2\log_2(2)

Then apply log rules and get rid off the logs.

Original post by Marcus2016

sorry to ask again but any hints for the following?

2 + log2(2x +1) =2 log2 y
x=22-y

RDKgames has suggested a nice method or you may like to rearrange the first equation to:

\displaystyle 2 \log_2 \left(y\right) - \log_2 \left(2x+1\right) = 2

and use the rule that

\displaystyle \log(a) - \log(b) = \log\left(\frac{a}{b}\right)

OP

Thanks again!!!

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