Question

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.

Answer 1

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)`