60log(1+r/12) = log1.8 textbook says to go log(1+r/12) = log1.8/60 i dont get the same answer. i figure you would have to (1+r/12) = log1.8/

$$60\times \log{(1+r/12)}=\log{1.8}$$

 

Divide both sides by 60

$$\log{(1+r/12)}=\frac{\log{1.8}}{60}$$

 

Assuming the log is to the base 10 then:

$$1+\frac{r}{12}=10^{\frac{\log{1.8}}{60}}$$

 

so:

$$r=12(10^{\frac{\log{1.8}}{60}}-1)$$

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