Question

1. a national tournament has 128 teams competing for the number 1 spot. after each

round, the amount of teams that move to the next round is halved. in which round will
there be less than 10 teams remaining? show or explain your reasoning.

1. it takes 3 years for ashley to double her initial collection of 150 pokémon cards. what is

the percentage that her collection increases yearly? show or explain your reasoning.
logarithms

1. ( log _{3} 6=? )
2. ( log _{10} 0.01=? )
3. ( log _{4} 1=? )
4. write the following exponential equations as logarithmic equations.

a. ( 3^{2}=9 quad
ightarrow )
b. ( 2^{-3}=\frac{1}{8} quad
ightarrow )

1. write the following logarithmic equations as exponential equations.

a. ( log _{5} 125=3 quad
ightarrow )
b. ( log _{2} 64=6 quad
ightarrow )

1. rewrite the equations as logarithms and then solve for the unknown exponent.

a. ( 4^{n}=256 )
b. ( 3^{p}=\frac{1}{27} )

Explanation:

1. $\log_{3}6 =?$

Step1: Use the change - of - base formula

The change - of - base formula is $\log_{a}b=\frac{\ln b}{\ln a}$. For $\log_{3}6$, we have $a = 3$ and $b = 6$. So, $\log_{3}6=\frac{\ln6}{\ln3}$.
Since $\ln6=\ln(2\times3)=\ln2+\ln3$, then $\log_{3}6=\frac{\ln2+\ln3}{\ln3}=1 + \frac{\ln2}{\ln3}$.
Using a calculator, $\ln2\approx0.693$ and $\ln3\approx1.099$. So, $\frac{\ln2}{\ln3}\approx0.631$. Then $\log_{3}6\approx1 + 0.631=1.631$.

2. $\log_{10}0.01 =?$

Step1: Use the definition of logarithms

We know that if $y = \log_{a}x$, then $x=a^{y}$. For $y=\log_{10}0.01$, we have $x = 0.01=\frac{1}{100}=10^{- 2}$ and $a = 10$.
By the definition $y=\log_{10}10^{-2}$. Since $\log_{a}a^{k}=k$ (where $a>0,a
eq1$), when $a = 10$ and $k=-2$, we get $\log_{10}0.01=-2$.

3. $\log_{4}1 =?$

Step1: Use the property of logarithms

We know that for any $a>0,a
eq1$, $\log_{a}1 = 0$. Because $a^{0}=1$ (by the zero - exponent rule: $x^{0}=1,x
eq0$). When $a = 4$, if $y=\log_{4}1$, then $4^{y}=1$. Since $4^{0}=1$, so $\log_{4}1 = 0$.

4.

a. $3^{2}=9$

Answer:

1. $\log_{3}6\approx1.631$
2. $\log_{10}0.01=-2$
3. $\log_{4}1 = 0$

4.

• a. $\log_{3}9 = 2$
• b. $\log_{2}\frac{1}{8}=-3$

5.

• a. $5^{3}=125$
• b. $2^{6}=64$

6.

• a. $n = 4$
• b. $p=-3$