lep checkpoint
complete the following problems. show all your work.
learning goal from lesson 15.1 and 16.1
i can solve problems involving logarithms and exponents because i
understand the inverse relationship between exponents and
logarithms.
how i did (circle one)
i got it! im still learning it.
1. which equation is equivalent to the following? (0.5 points total) (lesson 15.1)
a. (4^{x}=4096) b. (5^{-3}=\frac{1}{125})
a. (log _{4}x = 4096) a. (log _{5}\frac{1}{125}=-3)
b. (log _{4}4096 = x) b. (log _{\frac{1}{125}}-3 = 5)
c. (log _{x}4 = 4096) c. (log _{-3}5=\frac{1}{125})
d. (log _{4096}4 = x) d. (log _{5}-3=\frac{1}{125})
2. which equation is equivalent to the following? (0.5 points total) (lesson 15.1)
a. (log _{x}512 = 9) b. (log _{\frac{1}{6}}(\frac{1}{36}) = 2)
a. (9^{x}=512) a. ((2)^{\frac{1}{6}}=\frac{1}{36})
b. (x^{9}=512) b. ((\frac{1}{36})^{2}=\frac{1}{6})
c. (x^{512}=9) c. ((\frac{1}{6})^{\frac{1}{36}} = 2)
d. (512^{9}=x) d. ((\frac{1}{6})^{2}=\frac{1}{36})
3. let (f(x)=log _{5}x). what is the value of the following? (1 point each) (lesson 15.1)
1. (f(25)) 2. (f(\frac{1}{5})) 3. (f(sqrt{5}))

Question

Accepted Answer

1a.
# Explanation:
## Step1: Recall the logarithm - exponential relationship
If $a^{b}=c$, then $\log_{a}c = b$. For the equation $4^{x}=4096$, here $a = 4$, $b=x$ and $c = 4096$.
## Step2: Apply the relationship
By the formula $a^{b}=c\Leftrightarrow\log_{a}c = b$, we get $\log_{4}4096=x$

# Answer:
B. $\log_{4}4096 = x$

1b.
# Explanation:
## Step1: Recall the logarithm - exponential relationship
If $a^{b}=c$, then $\log_{a}c = b$. For the equation $5^{-3}=\frac{1}{125}$, here $a = 5$, $b=-3$ and $c=\frac{1}{125}$
## Step2: Apply the relationship
By the formula $a^{b}=c\Leftrightarrow\log_{a}c = b$, we get $\log_{5}\frac{1}{125}=-3$

# Answer:
A. $\log_{5}\frac{1}{125}=-3$

2a.
# Explanation:
## Step1: Recall the logarithm - exponential relationship
If $\log_{a}c = b$, then $a^{b}=c$. For the equation $\log_{x}512 = 9$, here $a=x$, $b = 9$ and $c = 512$
## Step2: Apply the relationship
By the formula $\log_{a}c=b\Leftrightarrow a^{b}=c$, we get $x^{9}=512$

# Answer:
B. $x^{9}=512$

2b.
# Explanation:
## Step1: Recall the logarithm - exponential relationship
If $\log_{a}c = b$, then $a^{b}=c$. For the equation $\log_{\frac{1}{6}}\frac{1}{36}=2$, here $a=\frac{1}{6}$, $b = 2$ and $c=\frac{1}{36}$
## Step2: Apply the relationship
By the formula $\log_{a}c=b\Leftrightarrow a^{b}=c$, we get $(\frac{1}{6})^{2}=\frac{1}{36}$

# Answer:
D. $(\frac{1}{6})^{2}=\frac{1}{36}$

3.
# Explanation:
## Step1: For $f(25)$
Since $f(x)=\log_{5}x$, and $25 = 5^{2}$. By the property $\log_{a}a^{k}=k$, when $a = 5$ and $k = 2$, $f(25)=\log_{5}25=\log_{5}5^{2}=2$
## Step2: For $f(\frac{1}{5})$
Since $\frac{1}{5}=5^{-1}$. By the property $\log_{a}a^{k}=k$, when $a = 5$ and $k=-1$, $f(\frac{1}{5})=\log_{5}\frac{1}{5}=\log_{5}5^{-1}=-1$
## Step3: For $f(\sqrt{5})$
Since $\sqrt{5}=5^{\frac{1}{2}}$. By the property $\log_{a}a^{k}=k$, when $a = 5$ and $k=\frac{1}{2}$, $f(\sqrt{5})=\log_{5}\sqrt{5}=\log_{5}5^{\frac{1}{2}}=\frac{1}{2}$

# Answer:
1. $2$
2. $-1$
3. $\frac{1}{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the logarithm - exponential relationship

Step2: Apply the relationship

Step1: Recall the logarithm - exponential relationship

Step2: Apply the relationship

Step1: Recall the logarithm - exponential relationship

Step2: Apply the relationship

Answer: