Linear regression doubling time

[yecko]

Homework Statement

see image

Relevant Equations

y=mx+c

y=mx+c
by z=log y axis
m=9-2.5 / 35 = 13/70
z=13/70 * (t-1983) + 2.5
log y = 13/70 * (t-1983) + 2.5 #

doubling time: t1=y1, t2=y2=2y1
log y1 = 13/70 * (t1-1983) + 2.5 ---{1}
log (2*y1) = 13/70 * (t2-1983) + 2.5 ---{2}
{2}-{1}: log2=13/70 * [(t2)-(t1)]
[(t2)-(t1)] = log2/(13/70) #

for log scale, can I log only y axis? for the constant c, is it 2.5 or 10^2.5?
is the doubling time calculated by simultaneous equation like my attempt?

Thank you.

 

[BvU]

y=mx+c
by z=log y axis
m=9-2.5 / 35 = 13/70
z=13/70 * (t-1983) + 2.5
log y = 13/70 * (t-1983) + 2.5 #

You are leaving us guessing what you did. A straight line would be $\ \log_{10} y = m x + c \$ or $y = 10^{c} \; \left (10^{x} \right )^m$

$m=9-2.5 / 35 \ne 13/70 \ \$ but $\ \ m= (9-2.5) / 35 = 13/70$

And yes, in 70 years $\log_{10}$ increases by 13 (actually: a little less !), so y becomes a factor 10¹³ bigger.

Doubling time is when $\ \ 13/70 * t = \log_{10} 2 \ \$. I wouldn't call that solving simultaneous equations...

for log scale, can I log only y axis?

What do you mean ? The verb logging means something else ¹³

 

[Mark44]

for log scale, can I log only y axis?

A graph can be semilog, such as log y vs x, or log-log, such as log y vs log x. The graph you showed is semilog, with t on the horizontal axis, and log(bits per sec) on the vertical axis.

Is that what you're asking?