Is My Textbook Incorrect On Explaining How To Solve Annual Interest Compounded Monthly? I doubt that the textbook solution is correct. If I have $\$100$ and put it into a bank with annual interest compounded monthly of $6\%$, how much money $y$ would I have after $t$ years?
The equation that the textbook provides is $y = 100(1+x)^{12t}$. The textbook states that for annual interest compounded monthly, the rate of $x$ equals $6\%/12$, which would equal $0.005$. But putting this into the equation with $t$ equaling $1$ for $1$ year, I get $106.1677812$ instead of the expected answer of simply $106$. Anyone know how to calculate $x$ to just get the correct exact answer of $106$? Is the equation incorrect as well?

An annual $6\%$ interest compounded monthly can mean either $0.005\%$ applied every month, or the interest is added monthly for a year's total of $6\%$.
In the second case (for $1$ year), we have the equation
$$(1+x)^{12}=1.06$$
In order to solve this, we use logarithms. Any base will do - I have used base $e$, the natural base.
Proceeding,
$$(1+x)^{12}=1.06$$
$$\ln\left((1+x)^{12}\right)=\ln(1.06)$$
$$12\ln(1+x)=0.05826890812397577552571835111851$$
$$\ln(1+x)=0.00485574234366464796047652925988$$
$$1+x=1.0048675505653430375411989455875$$
$$x=0.0048675505653430375411989455875$$
$x$ is the amount of interest added a month, to be written in the form of the question it needs to multiplied by $12$, giving an annual rate of $\approx 5.84\%$.
Another method is to take the $12^{th}$ root of both sides and subtract $1$ - this gives the same value for $x$.
