Converting
log39=2
32=9
log232
x=5
log2(7)+log2(4)
log228
log9(6-3w)=log9(-2w)
No Solution
Alexa invested $630 in an account paying an interest rate of 5.5% compounded quarterly. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $1,740?
t≈18.6
log61=0
60=1
log381
x=4
log5(100)-log5(25)
log5(4)
log4(m2)=log4(18-7m)
m= -9 and m=2
Joseph invested $85,000 in an account paying an interest rate of 1.6% compounded continuously. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $111,900?
t≈17.2
142=196
log14196=2
x=1/2 or 0.5
log(x2)-log(y3)
log(x2/y3)
log (2x)+log (x-5)=2
x=10
Jaxson is going to invest $180 and leave it in an account for 6 years. Assuming the interest is compounded daily, what interest rate, to the nearest tenth of a percent, would be required in order for Jaxson to end up with $240?
r≈4.8%
54=625
log5625=4
log168
x=3/4 or 0.75
(5)log(2)
log(32)
2log x-log 4=2
x=20
Mariana is going to invest $2,900 and leave it in an account for 7 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Mariana to end up with $3,300?
r≈1.85%
In(x-9)=32
e32=x-9
log243 27
x= 3/5 or .6
(1/3)log(8)
log(2)
2x=61
x= log2(61)
Owen is going to invest $550 and leave it in an account for 15 years. Assuming the interest is compounded monthly, what interest rate, to the nearest hundredth of a percent, would be required in order for Owen to end up with $1,280?
r≈5.64%