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

1. ## 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 2. 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 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 hope this speeds up getting an answer 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.$\displaystyle \log_{3} (x-1)=4$

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

Sorry but my minds gone blank about the rest.

4. Originally Posted by rotw121
1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years 2. Write in logarithmic form: 2^-2=1/4? 3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2? 4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4 Originally Posted by shinhidora [snip] 2.$\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 ..... 5. 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] $\displaystyle \log_3(x-1)=4$

$\displaystyle 3^4=x-1$

6. hey sorry guys i figured it out already but Thanks everyone for helping me out