Logs Unit

Converting

100

log₃9=2

3²=9 

200

log₆1=0

6⁰=1

300

14²=196

log₁₄196=2

400

5⁴=625

log₅625=4

500

In(x-9)=32

e³²=x-9

Evaluating

100

log₂32

x=5 

200

log₃81

x=4

300

log₁₀₀10

x=1/2 or 0.5

400

log₁₆8

x=3/4 or 0.75

500

log₂₄₃ 27

x= 3/5 or .6

Condensing

100

log₂(7)+log₂(4)

log₂28

200

log₅(100)-log₅(25)

log₅(4)

300

log(x²)-log(y³)

log(x²/y³)

400

(5)log(2)

log(32)

500

(1/3)log(8)

log(2)

Logarithmic

100

log₉(6-3w)=log₉(-2w)

No Solution

200

log₄(m2)=log₄(18-7m)

m= -9 and m=2

300

log (2x)+log (x-5)=2

x=10

400

2log x-log 4=2

x=20

500

2^x=61

x= log₂(61)

Application

100

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

200

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

300

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%

400

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%

500

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%