How to take logs of an equation?

Question

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

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

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 $y = m x + c$ $y=mx+c$ 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 $T = 2 π \sqrt{\frac{l}{g}}$ $T=2pisqrt(l/g)$

$log T = log 2 + log π + log \sqrt{l} - log \sqrt{g}$ $logT=log2+logpi+logsqrtl-logsqrtg$ ...............(A)

Now as $\sqrt{a} = a^{\frac{1}{2}}$ $sqrta=a^(1/2)$ , $log \sqrt{a} = {log a}^{\frac{1}{2}} = \frac{1}{2} log a$ $logsqrta=loga^(1/2)=1/2loga$

Also if we are using SI system of units $g = 9.81 \frac{m}{s^{2}}$ $g=9.81m/s^2$

and while $log 2 = 0.3010$ $log2=0.3010$ , $log π = 0.4971$ $logpi=0.4971$ and $log g = 0.9917$ $logg=0.9917$

Hence (A) becomes

$log T = 0.3010 + 0.4971 + \frac{1}{2} log l - \frac{1}{2} \times 0.9917$ $logT=0.3010+0.4971+1/2logl-1/2xx0.9917$

or $log T = 0.2996 + 0.5 log l$ $logT=0.2996+0.5logl$

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

So now it is in form $y = m x + c$ $y=mx+c$ , where $y = log T$ $y=logT$ , $m = \frac{1}{2}$ $m=1/2$ and $c = 0.2996$ $c=0.2996$ .

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

and then intercept should be $0.2996$ $0.2996$ and gradient would be $0.5$ $0.5$ . But intercept assumes $g = 9.81 \frac{m}{s^{2}}$ $g=9.81m/s^2$ .

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

Observe that intercept $c$ $c$ is actually $c = log (\frac{2 π}{\sqrt{g}})$ $c=log((2pi)/sqrtg)$

so once you get intercept $c$ $c$ , $c = log (\frac{2 π}{\sqrt{g}})$ $c=log((2pi)/sqrtg)$

and $\frac{2 π}{\sqrt{g}} = 10^{c}$ $(2pi)/sqrtg=10^c$ or $a n t i log c$ $antilogc$

i.e. $\sqrt{g} = \frac{2 π}{10^{c}}$ $sqrtg=(2pi)/10^c$ and $g = \frac{4 π^{2}}{10^{2 c}}$ $g=(4pi^2)/10^(2c)$