How do you solve ${{\left( 1+\left( \dfrac{0.064}{365} \right) \right)}^{365t}}=4$ ?

Hint: We are given equation as ${{\left( 1+\left( \dfrac{0.064}{365} \right) \right)}^{365t}}=4$ , we have to find the value if ‘t’, we look closely we get that we are given equation as exponential $a\left( {{b}^{t}} \right)$ , so to solve this exponential equation we will use the log function, we will apply log on both sides, then we will solve further using $\log \left( {{a}^{b}} \right)=b\log a$ , we will simplify term inside the bracket using BODMAS.

Complete step by step answer:
We are given equation as ${{\left( 1+\left( \dfrac{0.064}{365} \right) \right)}^{365t}}=4$ ,
We can see that our equation is in the form $a{{b}^{t}}=c$ form. So clearly we are given an exponential form of equation, to solve this problem and find value of ‘t’, we first solve the term inside the bracket as BODMAS says first we solve the bracket.
We have $\left( 1+\left( \dfrac{0.064}{365} \right) \right)$ , we first find $\dfrac{0.064}{365}$
The value of $\dfrac{0.064}{365}$ is $0.000175$ .
Now adding 1 to it so $\left( 1+\left( \dfrac{0.064}{365} \right) \right)$ becomes $1+0.000175$ .
So, $\Rightarrow \left( 1+\left( \dfrac{0.064}{365} \right) \right)=1.000175$ .
Now, we get –
$\Rightarrow {{\left( 1+\left( \dfrac{0.064}{365} \right) \right)}^{365t}}=4$
Will same as –
$\Rightarrow {{\left( 1.000175 \right)}^{365t}}=4$ .
Now we apply log on both sides.
$\log {{\left( 1.000175 \right)}^{365t}}=\log 4$ .
As we know $\log {{a}^{b}}=b\log a$ .
So, using this, we get –
$\Rightarrow 365t \log \left( 1.000175 \right)=2\log 2$ ( as $\log 4=\log {{2}^{2}}$ )
Now as we have to find the value of ‘t’.
So, we divide both sides by $365\log \left( 1.000175 \right)$ .
So, we get –
$\Rightarrow \dfrac{365t\log \left( 1.000175 \right)}{365\log \left( 1.000175 \right)}=\dfrac{2\log 2}{365\log \left( 1.000175 \right)}$
By simplifying, we get –
$\Rightarrow t=\dfrac{2\log 2}{365\log \left( 1.000175 \right)}$ .
By solving, we get –
$\Rightarrow t=21.705$ .

Note:
We need to know more about log, log has other properties as $\log \left( ab \right)=\log \left( a \right)+\log \left( b \right)$ , $\log \left( \dfrac{a}{b} \right)=\log a-\log b$ .
As we use $\log \left( {{a}^{b}} \right)=b\log a$ .
It came from the property $\log \left( ab \right)=\log a+\log b$ .
If we have $\log \left( {{a}^{b}} \right)$ then we can write it as $\log \left( a\times a \right)$ .
Then using first property we get –
$\log \left( {{a}^{b}} \right)=\log \left( a\times a \right)=\log a+\log b=2\log a$ .
Similarly if power is ‘b’ then $\log \left( {{a}^{b}} \right)=\log \left( a\times a\times a\times .......a \right)=\log a+\log a+......\log a=b\log a$ .