# Some math problems i cant seem to figure out? Part 3

• Jan 5th 2009, 07:09 PM
rotw121
Some math problems i cant seem to figure out? Part 3
1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years 2. Write in logarithmic form: 2^-2=1/4? 3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2? 4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4 • Jan 6th 2009, 08:55 AM shinhidora Quote: Originally Posted by rotw121 1. use the compound interest formula to determine the final value of the given amount:$1,000 at 8% compounded annually for 3 years

2. Write in logarithmic form: 2^-2=1/4?

3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2?

4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4

I myself can't really help you with logarithms :o I utterly, you know what, at them ;) but I've taken the liberty to put em in LaTex for ya ;) as I know some of the members don't like solving stuff that isn't put in a nice visual way :p hope this speeds up getting an answer :o and so I atleast help you in a certain way :)

2. $\displaystyle 2^{-2}=\frac{1}{4}$

3. $\displaystyle \log_{3} \frac{1}{9} = -2$

4.
a. $\displaystyle \log_{3} x=2$

b.$\displaystyle \log_{3} (x-1)=4$
• Jan 6th 2009, 09:05 AM
craig
Quote:

2. $\displaystyle 2^{-2}=\frac{1}{4}$
$\displaystyle \log_{2} {\frac{1}{4}}= -2$ :)

Sorry but my minds gone blank about the rest.
• Jan 6th 2009, 09:23 AM
mr fantastic
Quote:

Originally Posted by rotw121
1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years 2. Write in logarithmic form: 2^-2=1/4? 3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2? 4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4 Quote: Originally Posted by shinhidora [snip] 2.$\displaystyle 2^{-2}=\frac{1}{4}$3.$\displaystyle \log_{3} \frac{1}{9} = -2$4. a.$\displaystyle \log_{3} x=2$b.$\displaystyle \log_{3} (x-1)=4$Q3. and Q4. are done by applying the fact that$\displaystyle \log_A B = C \Rightarrow A^C = B$. For Q1, perhaps you could first tell us what the basic compound interest formula is ..... • Jan 6th 2009, 09:34 AM masters Quote: Originally Posted by rotw121 1. use the compound interest formula to determine the final value of the given amount:$1,000 at 8% compounded annually for 3 years

2. Write in logarithmic form: 2^-2=1/4?

3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2?

4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4

Hello rotw,

[1.] Compound interest formula:

$\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$

$\displaystyle A=1000\left(1+\frac{.08}{1}\right)^{3}$

Easy now....

[2.] $\displaystyle 2^{-2}=\frac{1}{4} \Longrightarrow \log_2 \frac{1}{4}=-2$

[3.] $\displaystyle \log_3 \frac{1}{9}=-2 \Longrightarrow 3^{-2}=\frac{1}{9}$

[4.1] $\displaystyle \log_3 x=2$

$\displaystyle x=3^2$

[4.2] $\displaystyle \log_3(x-1)=4$

$\displaystyle 3^4=x-1$
• Jan 7th 2009, 12:28 PM
rotw121
hey sorry guys i figured it out already but Thanks everyone for helping me out :)