How to take logs of an equation?

Question

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.

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Answer 1 · 2018-03-12T09:42:56

Please see below.

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 $g=(4pi^2)/10^(2c)$