# Problem

• March 16th 2013, 04:46 AM
endri
Problem
how can i solve this exercise? ( The value of an investment varies according to the formula V=A^t/12,where V is the value of the investment in pounds, A is the constant to be found and t is the time in years after the investment was made. i)If the investment was worth 8000 pounds after three years,how much the value of A to the nearest pound is? ii)How much the value of the investment after 10 years will be?

• March 16th 2013, 06:30 AM
Shakarri
Re: Problem
From
$V=A^{\frac{t}{12}}$

Raise both sides to the power of $\frac{12}{t}$

$V^{\frac{12}{t}}=A^{{\frac{t}{12}}^{\frac{12}{t}}}$

Remember the rules of indices. the indices ${\frac{t}{12}}$ and ${\frac{12}{t}}$ cancel to be just 1

$V^{\frac{12}{t}}=A^1$

You can now find A.
After you find the constant A you can do the second part easily by using the original equation at t=10
• March 16th 2013, 12:29 PM
endri
Re: Problem
shakari,,tell me what will be the answers with the data what i have written above..
• March 16th 2013, 12:43 PM
Shakarri
Re: Problem
• March 17th 2013, 02:00 AM
endri
Re: Problem
well if v=8000,then A=8000^4,in the first question, this is impossible..so plz tell me the answers