`(1+0.09/12)^(12t) = 3` Solve the equation accurate to three decimal places

`(1+0.09/12)^(12t) = 3`

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

Take the log of both sides of the equation.

`log_(10)((1+0.09/12)^(12t)) = log_10(3)`

With logarithms we know that;

`log(a^b) = bloga`

using that rule;

`log_(10)((1+0.09/12)^(12t)) = 12tlog_10(1+0.09/12)`

`12tlog_10(1+0.09/12) = log_10(3)`

`log_10(1+0.09/12) = log_10(1.0075) = 0.003245`

`log_10 (3) = 0.4771`

`12txx0.003245 = 0.4771`

`t = 0.4771/(12xx0.00325) = 12.252`

So the answer is t = 12.252

`(1+0.09/12)^(12t) = 3`

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

Take the log of both sides of the equation.

`log_(10)((1+0.09/12)^(12t)) = log_10(3)`

With logarithms we know that;

`log(a^b) = bloga`

using that rule;

`log_(10)((1+0.09/12)^(12t)) = 12tlog_10(1+0.09/12)`

`12tlog_10(1+0.09/12) = log_10(3)`

`log_10(1+0.09/12) = log_10(1.0075) = 0.003245`

`log_10 (3) = 0.4771`

`12txx0.003245 = 0.4771`

`t = 0.4771/(12xx0.00325) = 12.252`

So the answer is t = 12.252