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

Original post by maggiehodgson

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

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).

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).

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

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.

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?

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?

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?

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?

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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