Answer:
The correct option is d
Step-by-step explanation:
From the question we are told that
The equation is [tex]ln (SALARY) = 1 + 0.02 ln(EDU)[/tex]
generally
[tex]ln (e) = 1[/tex]
=> [tex]ln (SALARY) = ln (e) + 0.02 ln(EDU)[/tex]
Apply log rules
=> [tex]ln (SALARY) = ln[ e(EDU)^{0.02} ][/tex]
=> [tex]SALARY = e(EDU)^{0.02}[/tex]
When EDU is increase by 1% we have that
[tex]\frac{1}{100} * EDU + EDU[/tex]
=> [tex]1.01 EDU[/tex]
So the new SALARY will now be
[tex]SALARY'' = e(1.01EDU)^{0.02}[/tex]
The ratio by which the new SALARY will increase is obtain by dividing the value of new SALARY (SALARY'' ) with that of the former SALARY (SALARY) and this mathematically represented as
[tex] \frac{SALARY'' }{SALARY} = \frac{e*1.01 (EDU)^{0.02}}{e (EDU)^{0.02}}[/tex]
=> [tex]\frac{SALARY ''}{SALARY} = 1.0002[/tex]
Generally the percentage increase of salary is mathematically evaluated as
[tex]k= (\frac{SALARY''}{SALARY} - 1 ) * 100[/tex]
=> [tex]k= (1.0002 - 1 ) * 100[/tex]
=> [tex]k= 0.02 \%[/tex]
