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In v=ab^t. What does b represent. Logb is the gradient I think but what is b. Thanks

In v=ab^t. What does b represent. Logb is the gradient I think but what is b. Thanks
Reply 1
Original post by maggiehodgson
In v=ab^t. What does b represent. Logb is the gradient I think but what is b. Thanks
For the t-v relationship, its an exponential curve where b is the base and a is v(0) so the initial value. So if you increase t by 1, then y will scaled by being multiplied by b so
y(t+1) = b*y(t)

However, when you take logs you get
ln(v) = ln(a) + t*ln(b)
so the t-ln(v) relationship is linear with gradient ln(b) and "y"-intercept ln(a).
(edited 2 months ago)
Original post by mqb2766
For the t-v relationship, its an exponential curve where b is the base and a is v(0) so the initial value. So if you increase t by 1, then y will scaled by being multiplied by b so
y(t+1) = b*y(t)
However, when you take logs you get
ln(v) = ln(a) + t*ln(b)
so the t-ln(v) relationship is linear with gradient ln(b) and "y"-intercept ln(a).
I’m new to this version of tsr so perhaps I’ve already posted a reply. Apologies if I have.

Here is the actual maths question. I’m interested in b) ii) image.jpg


image.jpg
Reply 3
You have the linear equation (all log base 10)
log(y) = log(p) + t*log(q) = 4.8 + t*0.05
So
log(p) = 4.8
log(q) = 0.05
so
p = 10^4.8
q = 10^0.05 (must be slightly greater than 1, have you done 10^0.5?)
as its a "compound interest" type question (q-1)*100 would give the percentage "interest" each year. Each year, the price is multipled by q, so if it equals 1 it does not change. If its <1 the price decreases, if its >1 the price increases ... Similarly, taking logs, the log(v) graph would go up if log(q)>0 and down if log(q)<0 and be flat if log(q)=0.

Note for c) if you were unsure, you could get the 2010 price using the log(V) equation, then do 10^*. If you made an error on p and q, that part would not be affected.
(edited 2 months ago)
Thanks for pointing out my calculation error.
So q is either the compound growth factor if positive or the compound decay factor if negative? But log q would represent the gradient of the log graph?
Reply 5
Original post by maggiehodgson
Thanks for pointing out my calculation error.
So q is either the compound growth factor if positive or the compound decay factor if negative? But log q would represent the gradient of the log graph?
Yes, if q>1, then the expoential model grows and if q<1 it decays as each time increment youre multiplying the current amount by q. For an exponential, the base q is positive.

The exponential function/geometric sequence
v = pq^t
is equivalent to the log linear function
log(v) = log(p) + t*log(q)
and the growth / decay is determined by q and log(q). As 0 = log(1), the growth/decay boundary of the two models are equivaent. Its also worth emphasising that the original exponential function is the usual geometric sequence (t integer) with common ratio q.

Tbh, much of this is gcse and is roughly equivalent to the compound (exponential/geometric) interest and simple (linear) interest. The only proviso is that obviously compound and simple interest are not the same, but if you talk about log linear models as this question does, then theyre equivalent to the original exponential model.
(edited 2 months ago)

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