How to take logs of an equation?

Q: Take logs of both sides:
#T=2 pi sqrt frac{l}{g}#

My incomplete attempt at answering:
#log(T)=log(2pi sqrt frac {l}{g})#
#log(T)=log(2)+log(pi)+log(sqrtl)-log(sqrtg)#

I'm not sure if this is correct and I'm not sure how to remove the square roots.

Ultimately I'm trying to rearrange into the form #y=mx+c# so that I can input the gradient and intercept from a graph.

1 Answer
Mar 12, 2018

Answer:

Please see below.

Explanation:

As #T=2pisqrt(l/g)#

#logT=log2+logpi+logsqrtl-logsqrtg# ...............(A)

Now as #sqrta=a^(1/2)#, #logsqrta=loga^(1/2)=1/2loga#

Also if we are using SI system of units #g=9.81m/s^2#

and while #log2=0.3010#, #logpi=0.4971# and #logg=0.9917#

Hence (A) becomes

#logT=0.3010+0.4971+1/2logl-1/2xx0.9917#

or #logT=0.2996+0.5logl#

and here gradient is #0.5# and intercept is #0.2996#.

So now it is in form #y=mx+c#, where #y=logT#, #m=1/2# and #c=0.2996#.

Now you can draw a graph between #logT# and #logl#, where #T# is in seconds and #l# is in meters,

and then intercept should be #0.2996# and gradient would be #0.5#. But intercept assumes #g=9.81m/s^2#.

What if #g# is different? You can then get #g# from intercept.

Observe that intercept #c# is actually #c=log((2pi)/sqrtg)#

so once you get intercept #c#, #c=log((2pi)/sqrtg)#

and #(2pi)/sqrtg=10^c# or #antilogc#

i.e. #sqrtg=(2pi)/10^c# and #g=(4pi^2)/10^(2c)#