`(1+0.10/365)^(365t) = 2` Solve the equation accurate to three decimal places

`(1+0.1/365)^(365t) = 2`

In solving these kind of problems we need to use the logarithm.

Take the logarithm on both sides of the equation.

`log_(10)((1+0.1/365)^(365t)) = log_10(2)`

With logarithms we know that,

`log(a^b) = bloga`

using that rule;

`log_(10)((1+0.1/365)^(365t)) = 365tlog_10(1+0.1/365)`

`365tlog_10(1+0.1/365) = log_10(2)`

`log_10(1+0.1/365) = log_10(1.000274) = 0.000119`

`log_10 (2) = 0.3010`

`365txx0.000119 =0.3010`

`t = 0.3010/(365xx0.000119) = 6.931`

So the answer is t = 6.931

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