Actually 0.0312 would be the value of the log function to be exact, i.e., if you raise 10 to the power 0.0312 you would get 1+I as a result:
log (1+I) = 0.0312 can be re-written as 10^0.0312 = 1+I (the caret symbol ^ here meaning raised to the power, I could not find out how to write in superscript))
Since we're interested in finding out the value of I, we now solve for 1+I by simply performing the calculation 10^0.0312 on the calculator in scientific mode:
Input 10
Press x^y purple button in the center left side
Input 0.0312
You'll get a number 1.074481.... which can be rounded to 1.075, thus 1+I = 1.075. Subtracting 1 from each side of the equation gives the value of I = 0.075 (I'm sure this step was self explanatory).
log (1+I) = 0.0312 can be re-written as 10^0.0312 = 1+I (the caret symbol ^ here meaning raised to the power, I could not find out how to write in superscript))
Since we're interested in finding out the value of I, we now solve for 1+I by simply performing the calculation 10^0.0312 on the calculator in scientific mode:
Input 10
Press x^y purple button in the center left side
Input 0.0312
You'll get a number 1.074481.... which can be rounded to 1.075, thus 1+I = 1.075. Subtracting 1 from each side of the equation gives the value of I = 0.075 (I'm sure this step was self explanatory).
Excelsior BSB - MIS concentration (119 credits in the bank)