We are told that each year the value of the laptop is 75% of the value of the value of the previous year. This means that every year the current value of the laptop is multiplied by 0.75. Then if v is the original value of the laptop its value after t years is given by:
[tex]V(t)=v0.75^t[/tex]We need to find after which year V(t) is equal to 500 or less then we have V(t)≤500 and since the original value of the laptop was 4200 we have v=4200:
[tex]4200*0.75^t\leq500[/tex]We divide both sides by 4200:
[tex]\begin{gathered} \frac{4200\times0.75^t}{4200}\leqslant\frac{500}{4200} \\ 0.75^t\leq\frac{5}{42} \end{gathered}[/tex]Then we apply the logarithm to both sides:
[tex]\log_{10}(0.75^t)\leqslant\log_{10}(\frac{5}{42})[/tex]Then we use the property of logarithm regarding exponents:
[tex]\begin{gathered} \operatorname{\log}_{10}(0.75^{t})\leqslant\operatorname{\log}_{10}(\frac{5}{42}) \\ t\log_{10}(0.75)\leq\operatorname{\log}_{10}(\frac{5}{42}) \end{gathered}[/tex]And we divide both sides by the logarithm of 0.75 (we change the inequality symbol because log(0.75) is negative):
[tex]\begin{gathered} \frac{t\operatorname{\log}_{10}(0.75)}{\operatorname{\log}_{10}(0.75)}\ge\frac{\log_{10}(\frac{5}{42})}{\operatorname{\log}_{10}(0.75)} \\ t\ge\frac{\log_{10}(\frac{5}{42})}{\operatorname{\log}_{10}(0.75)} \end{gathered}[/tex]Then we get:
[tex]t\ge7.398[/tex]So the laptop's value is less than $500 after 7.398 years.
AnswerSince we are requested to write a whole number as the answer and the smallest whole number that is bigger than 7.398 is 8 we have that the answer is 8 years.
