Maths problem

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]

Originally Posted by 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

what's that ^ sign mean?

I don't remember seeing that in school

[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.

Originally Posted by Bishop

what's that ^ sign mean?

I don't remember seeing that in school

Yep Lord Nikon is correct

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 M]

Originally Posted by gary_580

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) ?

Originally Posted by 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

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]

Originally Posted by 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)

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...

Originally Posted by Paul M

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]

Originally Posted by gary_580

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]

Originally Posted by 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

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.