Maths problem

[Macca371]

Hi can anybody help me out here I'd appreciate it.

*Given that y = 2^x, express each of the following in terms of y.

a) 2^(x+1)

My answer was y^2 which is wrong :( I know the answer is 2y but I just don't know how to get to this answer. Can someone explain and also with this one...

b)2^(4x+3)


Answer is 8y^4.


First person who makes it click will get a green rep :)

[Bishop]

Hi can anybody help me out here I'd appreciate it.

*Given that y = 2^x, express each of the following in terms of y.

a) 2^(x+1)

My answer was y^2 which is wrong :( I know the answer is 2y but I just don't know how to get to this answer. Can someone explain and also with this one...

b)2^(4x+3)


Answer is 8y^4.


First person who makes it click will get a green rep :)

what's that ^ sign mean?

I don't remember seeing that in school

:dunce:

[Lord Nikon]

to the power of... x^2 = x squaed

[punky]

I'd PM BBKing nicely with it. He is the resident Boff. He did A-level maths I think.

[Macca371]

what's that ^ sign mean?

I don't remember seeing that in school

:dunce:

Yep Lord Nikon is correct

[gary_580]

not sure how your supposed to express this but
(its been 20yrs!!)

2^(x+1) = 2(2^x)

as y=2^x

then 2(2^x) = 2y

[Plonking]

lol,i got an N in as level maths..i tried to work it out, but i suck :( i'm ok with stats though if you have any problems with that...

[Paul]

2^(x+1) = 2(2^x)

as y=2^x

then 2(2^x) = 2y

How do you re-arrange 2^(x+1) to get 2(2^x) ?

[Macca371]

not sure how your supposed to express this but
(its been 20yrs!!)

2^(x+1) = 2(2^x)

as y=2^x

then 2(2^x) = 2y

Thankyou for your help I understand what you said but I'm still having trouble understanding the whole thing.

Never mind I'll ask my teacher tomorrow, that's what they are paid for at the end of the day.

[Theodoric]

Hi can anybody help me out here I'd appreciate it.

*Given that y = 2^x, express each of the following in terms of y.

a) 2^(x+1)

b)2^(4x+3)

The trick is to remember that (x^a).(x^b) = x^(a + b), in other words add the exponents of the 2 terms multiplied together.

Using this, you get:

a) 2^(x+1) = (2^x).2^1 = (2^x).2
As y = 2^x, then 2^(x + 1) = 2y

b) 2^(4x + 3) = (2^4x).2^3 = (2^x).(2^x).(2^x).(2^x).2^3 (as x + x + x + x = 4x)
Substituting y = 2^x gives y.y.y.y.2^3 = 8y^4

[Tristan]

Right. Let's put some numbers into this so you can see what's going on.

2^3=8, and 2^4=16

Now 8*16=128 (check it if you don't beleive me). So (2^3)*(2^4)=128.

But it just so happens that 128=2^7. What we've done is added the powers together -- so (2^3)*(2^4)=2^(3+4)=2^7=128

In the first example, we've got 2^(x+1). By doing the reverse of the above, you can see that 2^(x+1)=(2^x)*(2^1)=2*(2^x)=2y.

The second example is a little more complicated, so let me know if you "get" the above first.

I knew that Master's degree would come in handy sometime...

[gary_580]

How do you re-arrange 2^(x+1) to get 2(2^x) ?

because

2^(x+1)=(2^x).2^1

[Tristan]

I think we've covered it between us now! Shame that ordinary text makes this so hard to follow. I'm almost tempted to fire up LaTeX and do a nice pretty PDF!

[Theodoric]

because

2^(x+1)=(2^x).2^1
Yes, if you're working with exponents it often makes it clearer if you explicitly put in the exponent "1" where necessary; you can always remove it in the final answer if it appears there.

[nffc]

Hi can anybody help me out here I'd appreciate it.

*Given that y = 2^x, express each of the following in terms of y.

a) 2^(x+1)

My answer was y^2 which is wrong :( I know the answer is 2y but I just don't know how to get to this answer. Can someone explain and also with this one...

b)2^(4x+3)


Answer is 8y^4.


First person who makes it click will get a green rep :)
You can also log them, if there was an equality which there isn't, and if you didn't know y=2^x. This is useful if you have a problem y = x ^n and need to solve for n... [note- this is not relevant to the current problem which has been solved, but may be useful for the future]

a) z=2^(x+1), logging both sides gives you ln z = ln (2^(x+1))

if you log an index you can bring the index down...

ln z = (x+1) ln 2

Rearrange to get (ln z / ln 2) -1 = x

for example. :)

Maths is ****e.

[Paul]

because

2^(x+1)=(2^x).2^1Thanks, between you and Tristan I see it now. :D

[basa]

OMG there's some bright b*st*rds on this forum :D

My missus is struggling with GCSE Intermediate Maths atm and whilst I easily got my GCE Maths 'O' Level back in '65 and went on to do Maths at Higher National level and (nearly :confused: ) mastered Integral and Differential Calculus this lot stumped me. :rolleyes:

[Tristan]

OMG there's some bright b*st*rds on this forum :D

My missus is struggling with GCSE Intermediate Maths atm and whilst I easily got my GCE Maths 'O' Level back in '65 and went on to do Maths at Higher National level and (nearly :confused: ) mastered Integral and Differential Calculus this lot stumped me. :rolleyes:

If you mastered integral calculus, this should be a breeze by comparison. Integration only just makes it onto (basic) A-Level syllabus these days.

The question here is, I think, probably higher GCSE level. It's the kind of stuff you have to know before you start dealing with logarithms. Interestingly, my dad tells me he did logs on his old O-Level maths paper.


Tristan: Bright B*stard in Chief ;)

[downquark1]

Integration only just makes it onto (basic) A-Level syllabus these days. Although I understand that they are changing it again, integration rears its head in the pure 1 module which is generally used for AS level (1st year of an A level)

[Macca371]

Thankyou Theodoric, Tristan, nffc, and Gary.

I'll rep all four :)

[nffc]

Although I understand that they are changing it again, integration rears its head in the pure 1 module which is generally used for AS level (1st year of an A level)
Were both in P1 when I did it, banely simple once you learn the rules, and it's the processing that's more difficult.

[homealone]

If you mastered integral calculus, this should be a breeze by comparison. Integration only just makes it onto (basic) A-Level syllabus these days.

The question here is, I think, probably higher GCSE level. It's the kind of stuff you have to know before you start dealing with logarithms. Interestingly, my dad tells me he did logs on his old O-Level maths paper.


Tristan: Bright B*stard in Chief ;)

'did logs' - you had to use logs to do some of the questions, the logs had to be looked up in a book of tables that included the values for logs to base 10, logs to base e and the values for sine/cosine/tangent etc, of angles, for the trig questions. :)

[nffc]

'did logs' - you had to use logs to do some of the questions, the logs had to be looked up in a book of tables that included the values for logs to base 10, logs to base e and the values for sine/cosine/tangent etc, of angles, for the trig questions. :)
yeh, presumably in the pre-scientific calculator era! ;) :D

We were allowed graphic calculators in exams :)

BTW- I have AS-level maths and at uni we've been taught to probably just over A-level standard in pure maths. Maths still sucks though :(

[homealone]

yeh, presumably in the pre-scientific calculator era! ;) :D

We were allowed graphic calculators in exams :)

BTW- I have AS-level maths and at uni we've been taught to probably just over A-level standard in pure maths. Maths still sucks though :(

I grew to love it, rather than, it loved me :)

- things like statistics, physics, chemistry & biology 'rely' on maths ;)

[nffc]

I grew to love it, rather than, it loved me :)

- things like statistics, physics, chemistry & biology 'rely' on maths ;) Yeah I'm a chemist, and i have a-level physics and biology so I'm well aware of that, it still sucks though :(

[MetaWraith]

Coming into this thread a bit late, I have this observation to make.
There is no need to involve logs, just simple algebra and a working knowlege of powers.

given that the caret symbol ^ is used to denote "raised to the power of" just remember that powers can be decomposed thusly
P^(a+b) = (P^a) * (P^b) thus 2^(3+2) = (2^3) * (2^2) = 8*4 = 32
P^(m*n) = (P^m)^n thus 2^(3*2) = (2^3)^2 = 8^2 = 64

thus for 1)
2^(x+1) = (2^1) * (2^x) = 2 * y = 2y

and for 2)
2^(4x+3) = (2^3) * (2^4x) = 8 * (2^x)^4 = 8y^4

couldnt be simpler }:¬)

[homealone]

Yeah I'm a chemist, and i have a-level physics and biology so I'm well aware of that, it still sucks though :(

lol - seeking 'equilibrium', as we speak ;)

[MovedGoalPosts]

couldnt be simpler }:¬)

I'll take your word for it ;)

I'm amazed I thought I was reasonably good at maths all those years ago when I was a youngster at school. It just goes to show how much you forget when you don't use it. :dunce:

[gary_580]

I'll take your word for it ;)

I'm amazed I thought I was reasonably good at maths all those years ago when I was a youngster at school. It just goes to show how much you forget when you don't use it. :dunce:


or just shows how pointless it is :D

[MetaWraith]

lol - seeking 'equilibrium', as we speak ;)
Is that a pint in each hand ?

[homealone]

Is that a pint in each hand ?

maybe a glass of wine - bulgarian, cabernet :)

[nffc]

Coming into this thread a bit late, I have this observation to make.
There is no need to involve logs, just simple algebra and a working knowlege of powers.

given that the caret symbol ^ is used to denote "raised to the power of" just remember that powers can be decomposed thusly
P^(a+b) = (P^a) * (P^b) thus 2^(3+2) = (2^3) * (2^2) = 8*4 = 32
P^(m*n) = (P^m)^n thus 2^(3*2) = (2^3)^2 = 8^2 = 64

thus for 1)
2^(x+1) = (2^1) * (2^x) = 2 * y = 2y

and for 2)
2^(4x+3) = (2^3) * (2^4x) = 8 * (2^x)^4 = 8y^4

couldnt be simpler }:¬)
Yeah I knew (and said!) that, in fact you can't use logs as there's no equality, but was just pointing out an aside!

Incidentally the laws of logs and indices are the same anyway...

- a^b. a^c = a ^(b+c) is the same as ln (xy)=ln x + ln y
- a^b/ a^c = a^(b=c) is the same as ln (x/y)=ln x - ln y
- (a^b)^c = a^(bc) is the same as ln (x^y)= y ln x

so it's the same thing...

But where there's no equality you can't use logs as you have to do the same to both sides of the equality- as there's not two sides to the equality as there's no equality, you can't log.

It was purely for information!

[basa]

If you mastered integral calculus, this should be a breeze by comparison. Integration only just makes it onto (basic) A-Level syllabus these days.

The question here is, I think, probably higher GCSE level. It's the kind of stuff you have to know before you start dealing with logarithms. Interestingly, my dad tells me he did logs on his old O-Level maths paper.


Tristan: Bright B*stard in Chief ;)

You'd be right only this was some 35 years ago !! :o and I've forgotten most of it ! :shocked:

I did do logs (and sines, cosines, tangents etc) from a little A5 sized book - I think I might even have that book in the loft !

All that stuff just passes me by now ! :dunce: (Thanks god !) :disturbd:

[BBKing]

Bit late to this one.

Did we get my two favourite log related facts in?

1) log x / log y = log to the base y of x (the logs can be to any base you like). Makes it very easy to find the log to the base anything of anything, given a working log function (or log tables for the ancients).

2) ln 5 (log to the base e of 5) is near as dammit the ratio of miles to kilometres (1.605...)

Integration only just makes it onto (basic) A-Level syllabus these days.

And they say it isn't dumbed-down? When I was at uni we went straight for calculus from day one (structural engineering needs it). Those on the course who'd got less than a C at A-level had to do a maths refresher course for the first year. You had to know your calculus.

[nffc]

Bit late to this one.

Did we get my two favourite log related facts in?

1) log x / log y = log to the base y of x (the logs can be to any base you like). Makes it very easy to find the log to the base anything of anything, given a working log function (or log tables for the ancients).

2) ln 5 (log to the base e of 5) is near as dammit the ratio of miles to kilometres (1.605...)
What's ln (fire) then? :)

And they say it isn't dumbed-down? When I was at uni we went straight for calculus from day one (structural engineering needs it). Those on the course who'd got less than a C at A-level had to do a maths refresher course for the first year. You had to know your calculus.
Nah, we did calculus in AS...