SOLUTION: Solve: 1) 5^2X^2^+^3X=25 2) 3^-X=7^X+^2 3) 250000=100000(1+.06/12)^12(t)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve: 1) 5^2X^2^+^3X=25 2) 3^-X=7^X+^2 3) 250000=100000(1+.06/12)^12(t)      Log On


   

Question 1139283: Solve:
1) 5^2X^2^+^3X=25

2) 3^-X=7^X+^2

3) 250000=100000(1+.06/12)^12(t)
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


1) 5^2X^2^+^3X=25

The sequence of characters "X^2+^" has no meaning.
The sequence of characters "+^3" has no meaning.

2) 3^-X=7^X+^2

The sequence of characters "+^2" has no meaning.

3) 250000=100000(1+.06/12)^12(t)

The equation as you show it is solved by ordinary arithmetic....

250000=100000%281%2B.06%2F12%29%5E12%28t%29

The value of t is 250000, divided by the product of 100000 and (1+.06/12)^12.

Surely you meant

3) 250000=100000(1+.06/12)^(12t)

This is now an equation for finding the number of years t that it takes 100000 to increase to 250000 at 6% compounded monthly.

250000=100000%281%2B.06%2F12%29%5E%2812t%29
2.5+=+%281.005%29%5E%2812t%29

The variable is in the exponent; you need to take logs of both sides.

log%28%282.5%29%29+=+%2812t%29%2Alog%28%281.005%29%29
12t+=+log%28%282.5%29%29%2Flog%28%281.005%29%29

Use a calculator....