Math / Algebra

QuestionLESSON 5.3 Worksheet A: (Topic 2.3 ) Directions: Convert the following equation
4. $y=e^{x-1}$
1. $2^{3}=8$
2. $10^{3}=1000$
Directions: Convert the following equations from logarithmic (log) form to exponential form.
5. $\log _{4} 1=0$
6. $\log \frac{1}{100}=-2$
7. $\log _{16} y=\frac{1}{2}$
8. $\ln x=4$
Directions: Evaluate the following expressions without a calculator.
9. $\log _{3} 9=2$
10. $\log _{6} \sqrt{36}=2$
11. $\ln 1$
12. $\log 10$
13. $\log _{4} \frac{1}{16}$
14. $\log _{2} 8$
15. $\ln e^{7}$
16. $\ln \frac{1}{e^{5}}$
17. $\log _{25} 5$
18. $\log _{16} 2$
19. $\log _{6} \sqrt{6}$
20. $\log _{9} \frac{1}{3}$

Studdy Solution

STEP 1

What is this asking? We're going to be rockstars at switching between logs and exponents, and then we'll become lightning-fast at calculating logs without a calculator! Watch out! Remember that logs and exponents are just two sides of the same coin.
Don't let the different notations trip you up!

STEP 2

1. Convert exponential equations to logarithmic form.
2. Convert logarithmic equations to exponential form.
3. Evaluate logarithmic expressions.

STEP 3

Let's take $2^3 = 8$ .
This tells us that $2$ raised to the power of $3$ gives us $8$ .
To write this in log form, we ask ourselves, "What power do we need to raise $2$ to in order to get $8$ ?".
The answer is $3$ !
So, we write $\log_{2} 8 = 3$ .
The base of the exponent becomes the base of the log!

STEP 4

Now let's try $10^3 = 1000$ .
What power do we raise $10$ to, to get $1000$ ?
It's $3$ !
So, $\log_{10} 1000 = 3$ .
When the base is $10$ , we sometimes just write $\log$ without the little number.
So, we could also write $\log 1000 = 3$ .

STEP 5

Lastly, we have $y = e^{x-1}$ .
Here, $e$ is the base.
We're asking, "What power do we raise $e$ to, to get $y$ ?".
The answer is $x-1$ !
So, we write $\ln y = x-1$ . $\ln$ is a special shorthand for $\log_{e}$ , called the natural logarithm.

STEP 6

Let's look at $\log_{4} 1 = 0$ .
This is asking, "What power do we raise $4$ to, to get $1$ ?".
The answer is $0$ !
Any number (except $0$ ) raised to the power of $0$ is $1$ .
So, we write $4^0 = 1$ .

STEP 7

Now for $\log \frac{1}{100} = -2$ .
Remember, when the base isn't written, it's $10$ .
So, we're asking, "What power do we raise $10$ to, to get $\frac{1}{100}$ ?" The answer is $-2$ !
So, $10^{-2} = \frac{1}{100}$ .

STEP 8

Let's try $\log_{16} y = \frac{1}{2}$ .
This means $16$ raised to the power of $\frac{1}{2}$ gives us $y$ .
So, $16^{\frac{1}{2}} = y$ .
Remember, a fractional exponent is just a root! $16^{\frac{1}{2}}$ is the same as $\sqrt{16}$ , which is $4$ .
So, $y=4$ .

STEP 9

Finally, we have $\ln x = 4$ .
Remember, $\ln$ means $\log_{e}$ .
So, this means $e^4 = x$ .

STEP 10

$\log_{3} 9$ asks, "What power do we raise $3$ to, to get $9$ ?" It's $2$ , since $3^2 = 9$ .
So, $\log_{3} 9 = 2$ .

STEP 11

$\log_{6} \sqrt{36}$ . $\sqrt{36}$ is just $6$ .
So, $\log_{6} 6$ .
What power do we raise $6$ to, to get $6$ ?
It's $1$ !
So, $\log_{6} \sqrt{36} = 1$ .

STEP 12

$\ln 1$ asks, "What power do we raise $e$ to, to get $1$ ?" Any number (except $0$ ) raised to the power of $0$ is $1$ .
So, $\ln 1 = 0$ .

STEP 13

$\log 10$ has an invisible base of $10$ .
What power do we raise $10$ to, to get $10$ ?
It's $1$ !
So, $\log 10 = 1$ .

STEP 14

$\log_{4} \frac{1}{16}$ .
We can rewrite $\frac{1}{16}$ as $16^{-1}$ , which is $(4^2)^{-1}$ or $4^{-2}$ .
So, $\log_{4} 4^{-2} = -2$ .

STEP 15

$\log_{2} 8$ .
Since $8 = 2^3$ , we have $\log_{2} 2^3 = 3$ .

STEP 16

$\ln e^7 = 7$ .
Because the base of the logarithm and the base of the exponent are both $e$ , the result is simply the exponent.

STEP 17

$\ln \frac{1}{e^5} = \ln e^{-5} = -5$ .

STEP 18

$\log_{25} 5$ .
Since $5$ is the square root of $25$ , which is $25^{\frac{1}{2}}$ , we have $\log_{25} 25^{\frac{1}{2}} = \frac{1}{2}$ .

STEP 19

$\log_{16} 2$ .
Since $2$ is the fourth root of $16$ , which is $16^{\frac{1}{4}}$ , we have $\log_{16} 16^{\frac{1}{4}} = \frac{1}{4}$ .

STEP 20

$\log_{6} \sqrt{6} = \log_{6} 6^{\frac{1}{2}} = \frac{1}{2}$ .

STEP 21

$\log_{9} \frac{1}{3}$ .
We can rewrite this as $\log_{9} 3^{-1}$ .
Since $9 = 3^2$ , we're looking for $\log_{3^2} 3^{-1}$ .
This is equal to $-\frac{1}{2}$ .

STEP 22

We successfully converted between exponential and logarithmic forms, and evaluated a bunch of logarithmic expressions like pros!

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Studdy Solution

STEP 1

What is this asking? We're going to be rockstars at switching between logs and exponents, and then we'll become lightning-fast at calculating logs without a calculator! Watch out! Remember that logs and exponents are just two sides of the same coin.Don't let the different notations trip you up!

STEP 2

1. Convert exponential equations to logarithmic form.2. Convert logarithmic equations to exponential form.3. Evaluate logarithmic expressions.

STEP 3

STEP 4

STEP 5

Lastly, we have y=ex−1y = e^{x-1}y=ex−1.Here, eee is the **base**.We're asking, "What power do we raise eee to, to get yyy?".The answer is x−1x-1x−1!So, we write ln⁡y=x−1\ln y = x-1lny=x−1. ln⁡\lnln is a special shorthand for log⁡e\log_{e}loge​, called the **natural logarithm**.

STEP 6

Let's look at log⁡41=0\log_{4} 1 = 0log4​1=0.This is asking, "What power do we raise 444 to, to get 111?".The answer is 000!Any number (except 000) raised to the power of 000 is 111.So, we write 40=14^0 = 140=1.

STEP 7

Now for log⁡1100=−2\log \frac{1}{100} = -2log1001​=−2.Remember, when the base isn't written, it's 101010.So, we're asking, "What power do we raise 101010 to, to get 1100\frac{1}{100}1001​?" The answer is −2-2−2!So, 10−2=110010^{-2} = \frac{1}{100}10−2=1001​.

STEP 8

STEP 9

Finally, we have ln⁡x=4\ln x = 4lnx=4.Remember, ln⁡\lnln means log⁡e\log_{e}loge​.So, this means e4=xe^4 = xe4=x.

STEP 10

log⁡39\log_{3} 9log3​9 asks, "What power do we raise 333 to, to get 999?" It's 222, since 32=93^2 = 932=9.So, log⁡39=2\log_{3} 9 = 2log3​9=2.

STEP 11

log⁡636\log_{6} \sqrt{36}log6​36​. 36\sqrt{36}36​ is just 666.So, log⁡66\log_{6} 6log6​6.What power do we raise 666 to, to get 666?It's 111!So, log⁡636=1\log_{6} \sqrt{36} = 1log6​36​=1.

STEP 12

ln⁡1\ln 1ln1 asks, "What power do we raise eee to, to get 111?" Any number (except 000) raised to the power of 000 is 111.So, ln⁡1=0\ln 1 = 0ln1=0.

STEP 13

log⁡10\log 10log10 has an invisible base of 101010.What power do we raise 101010 to, to get 101010?It's 111!So, log⁡10=1\log 10 = 1log10=1.

STEP 14

log⁡4116\log_{4} \frac{1}{16}log4​161​.We can rewrite 116\frac{1}{16}161​ as 16−116^{-1}16−1, which is (42)−1(4^2)^{-1}(42)−1 or 4−24^{-2}4−2.So, log⁡44−2=−2\log_{4} 4^{-2} = -2log4​4−2=−2.

STEP 15

log⁡28\log_{2} 8log2​8.Since 8=238 = 2^38=23, we have log⁡223=3\log_{2} 2^3 = 3log2​23=3.

STEP 16

ln⁡e7=7\ln e^7 = 7lne7=7.Because the base of the logarithm and the base of the exponent are both eee, the result is simply the exponent.

STEP 17

ln⁡1e5=ln⁡e−5=−5\ln \frac{1}{e^5} = \ln e^{-5} = -5lne51​=lne−5=−5.

STEP 18

log⁡255\log_{25} 5log25​5.Since 555 is the square root of 252525, which is 251225^{\frac{1}{2}}2521​, we have log⁡252512=12\log_{25} 25^{\frac{1}{2}} = \frac{1}{2}log25​2521​=21​.

STEP 19

log⁡162\log_{16} 2log16​2.Since 222 is the fourth root of 161616, which is 161416^{\frac{1}{4}}1641​, we have log⁡161614=14\log_{16} 16^{\frac{1}{4}} = \frac{1}{4}log16​1641​=41​.

STEP 20

log⁡66=log⁡6612=12\log_{6} \sqrt{6} = \log_{6} 6^{\frac{1}{2}} = \frac{1}{2}log6​6​=log6​621​=21​.

STEP 21

log⁡913\log_{9} \frac{1}{3}log9​31​.We can rewrite this as log⁡93−1\log_{9} 3^{-1}log9​3−1.Since 9=329 = 3^29=32, we're looking for log⁡323−1\log_{3^2} 3^{-1}log32​3−1.This is equal to −12-\frac{1}{2}−21​.

STEP 22

We successfully converted between exponential and logarithmic forms, and evaluated a bunch of logarithmic expressions like pros!

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What is this asking? We're going to be rockstars at switching between logs and exponents, and then we'll become lightning-fast at calculating logs without a calculator! Watch out! Remember that logs and exponents are just two sides of the same coin.
Don't let the different notations trip you up!

1. Convert exponential equations to logarithmic form.
2. Convert logarithmic equations to exponential form.
3. Evaluate logarithmic expressions.

Lastly, we have $y = e^{x-1}$ .
Here, $e$ is the base.
We're asking, "What power do we raise $e$ to, to get $y$ ?".
The answer is $x-1$ !
So, we write $\ln y = x-1$ . $\ln$ is a special shorthand for $\log_{e}$ , called the natural logarithm.

Let's look at $\log_{4} 1 = 0$ .
This is asking, "What power do we raise $4$ to, to get $1$ ?".
The answer is $0$ !
Any number (except $0$ ) raised to the power of $0$ is $1$ .
So, we write $4^0 = 1$ .

Now for $\log \frac{1}{100} = -2$ .
Remember, when the base isn't written, it's $10$ .
So, we're asking, "What power do we raise $10$ to, to get $\frac{1}{100}$ ?" The answer is $-2$ !
So, $10^{-2} = \frac{1}{100}$ .

Finally, we have $\ln x = 4$ .
Remember, $\ln$ means $\log_{e}$ .
So, this means $e^4 = x$ .

$\log_{3} 9$ asks, "What power do we raise $3$ to, to get $9$ ?" It's $2$ , since $3^2 = 9$ .
So, $\log_{3} 9 = 2$ .

$\log_{6} \sqrt{36}$ . $\sqrt{36}$ is just $6$ .
So, $\log_{6} 6$ .
What power do we raise $6$ to, to get $6$ ?
It's $1$ !
So, $\log_{6} \sqrt{36} = 1$ .

$\ln 1$ asks, "What power do we raise $e$ to, to get $1$ ?" Any number (except $0$ ) raised to the power of $0$ is $1$ .
So, $\ln 1 = 0$ .

$\log 10$ has an invisible base of $10$ .
What power do we raise $10$ to, to get $10$ ?
It's $1$ !
So, $\log 10 = 1$ .

$\log_{4} \frac{1}{16}$ .
We can rewrite $\frac{1}{16}$ as $16^{-1}$ , which is $(4^2)^{-1}$ or $4^{-2}$ .
So, $\log_{4} 4^{-2} = -2$ .

$\log_{2} 8$ .
Since $8 = 2^3$ , we have $\log_{2} 2^3 = 3$ .

$\ln e^7 = 7$ .
Because the base of the logarithm and the base of the exponent are both $e$ , the result is simply the exponent.

$\ln \frac{1}{e^5} = \ln e^{-5} = -5$ .

$\log_{25} 5$ .
Since $5$ is the square root of $25$ , which is $25^{\frac{1}{2}}$ , we have $\log_{25} 25^{\frac{1}{2}} = \frac{1}{2}$ .

$\log_{16} 2$ .
Since $2$ is the fourth root of $16$ , which is $16^{\frac{1}{4}}$ , we have $\log_{16} 16^{\frac{1}{4}} = \frac{1}{4}$ .

$\log_{6} \sqrt{6} = \log_{6} 6^{\frac{1}{2}} = \frac{1}{2}$ .

$\log_{9} \frac{1}{3}$ .
We can rewrite this as $\log_{9} 3^{-1}$ .
Since $9 = 3^2$ , we're looking for $\log_{3^2} 3^{-1}$ .
This is equal to $-\frac{1}{2}$ .