# 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: $\log \left(T\right) = \log \left(2 \pi \sqrt{\frac{l}{g}}\right)$ $\log \left(T\right) = \log \left(2\right) + \log \left(\pi\right) + \log \left(\sqrt{l}\right) - \log \left(\sqrt{g}\right)$ 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 $y = m x + c$ so that I can input the gradient and intercept from a graph.

Mar 12, 2018

#### Explanation:

As $T = 2 \pi \sqrt{\frac{l}{g}}$

$\log T = \log 2 + \log \pi + \log \sqrt{l} - \log \sqrt{g}$ ...............(A)

Now as $\sqrt{a} = {a}^{\frac{1}{2}}$, $\log \sqrt{a} = \log {a}^{\frac{1}{2}} = \frac{1}{2} \log a$

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

and while $\log 2 = 0.3010$, $\log \pi = 0.4971$ and $\log g = 0.9917$

Hence (A) becomes

$\log T = 0.3010 + 0.4971 + \frac{1}{2} \log l - \frac{1}{2} \times 0.9917$

or $\log T = 0.2996 + 0.5 \log l$

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

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

Now you can draw a graph between $\log T$ and $\log l$, 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 $g = 9.81 \frac{m}{s} ^ 2$.

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

Observe that intercept $c$ is actually $c = \log \left(\frac{2 \pi}{\sqrt{g}}\right)$

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

and $\frac{2 \pi}{\sqrt{g}} = {10}^{c}$ or $a n t i \log c$

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