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During a 5-year period of constant inflation, the value of a $113,000 property increases according to the equation v=113,000e0.04t dollars. In how many years will the value of this building be double its current value?

During A 5year Period Of Constant Inflation The Value Of A 113000 Property Increases According To The Equation V113000e004t Dollars In How Many Years Will The V class=

Sagot :

Answer:

17.33 years.

Explanation:

The value of the property increases according to the equation:

[tex]v=113,000e^{0.04t}[/tex]

When the building is double its current value:

[tex]\begin{gathered} v=2\times113,000. \\ \implies2\times113,000=113,000e^{0.04t} \end{gathered}[/tex]

We want to solve for t:

Divide both sides by 113,000

[tex]\begin{gathered} \frac{2\times113,000}{113,000}=\frac{113,000e^{0.04t}}{113,000} \\ 2=e^{0.04t} \end{gathered}[/tex]

Take the natural logarithm (ln) of both sides:

[tex]\begin{gathered} \ln (2)=\ln (e^{0.04t}) \\ \ln (2)=0.04t \end{gathered}[/tex]

Finally, divide both sides by 0.04:

[tex]\begin{gathered} \frac{\ln (2)}{0.04}=\frac{0.04t}{0.04} \\ t=17.33 \end{gathered}[/tex]

In 17.33 years, the value of the building will double its current value.