MATEMATİK I

2³*2^x*5*5^x=4000.

40*10^x=4000

10^x=100

so x=2

1000*1.2*1.2*1.2=1728 TL

So the interest earning is 1728-1000=728

log_a(xy) = log_ax + log_ay
log_a(x/y) = log_ax - log_ay

(e^x - 1)*(e^x + 6)=0

so either e^x - 1=0 or e^x + 6=0

If e^x - 1=0 then e^x = 1 which means x=0.

If e^x + 6=0 then e^x = -6 which can not be the solution because it is not in domain of the natural nase function e^x . 

Thus x=0.

(e^x - 4)(e^x - 3)=0

Thus the equation has two roots:

for e^x - 4=0, e^x = 4 so x=ln4

for e^x - 3=0, e^x = 3 so x=ln3

Since we know that ln1=0,

ln 0.5=0-a=-a

Thus P(t=2)=P(0)e^2k . So:

900=100e^2k

9=e^2k=(e^k)²

3²=(e^k)²

3=e^k

The population after 10 hours will be:

P(t=10)=100e^10k=100*(e^k)¹⁰ =100*3¹⁰

0.5 Q= Q.e^400k 

Thus 0.5 = e^400k so ln 0.5=ln e^400k =400k so k=ln 0.5 /400

Q(t=600)=1000*e^600k

let e^600k = b,

then 600k = ln b and since we found that k = ln 0.5/400

600* ln 0.5 /400= ln b=1.5 ln 0.5

thus b=(0.5)^1.5 = 0.3535

Then Q(t=600)=1000*0.3535=353.5 gr=354 gr (approximately)