Math / Algebra

QuestionSolve by using the equivalence property of logarithmic expressions, and write the solution set.
1. $\ln \left(x^{2}-66\right)=\ln (-5 x)$
Solve by using exponential form, and write the solution set.
2. $8 \log (3 x+28)-5=11$

Studdy Solution

STEP 1

What is this asking? We need to solve two logarithmic equations, one using the equivalence property and the other using exponential form. Watch out! Remember to check if your solutions are valid by plugging them back into the original logarithmic equations.
Logarithms are only defined for positive arguments!

STEP 2

1. Solve using equivalence property
2. Solve using exponential form

STEP 3

Alright, let's tackle the first equation: $\ln(x^2 - 66) = \ln(-5x)$ .
Since the logarithms are equal, their arguments must be equal too!
This is the equivalence property!

STEP 4

So, we set $x^2 - 66 = -5x$ .
Now, we've got a quadratic equation.
Let's rearrange it to $x^2 + 5x - 66 = 0$ .

STEP 5

Time to factor!
We're looking for two numbers that multiply to $-66$ and add up to $5$ .
Those numbers are $11$ and $-6$ .
So, we can rewrite the equation as $(x + 11)(x - 6) = 0$ .

STEP 6

This gives us two potential solutions: $x = -11$ and $x = 6$ .
But hold on!
We need to check if these solutions are valid.

STEP 7

Let's check $x = -11$ .
Plugging it back into the original equation, we get $\ln((-11)^2 - 66) = \ln(-5 \cdot (-11))$ , which simplifies to $\ln(121 - 66) = \ln(55)$ , or $\ln(55) = \ln(55)$ .
This works!

STEP 8

Now, let's check $x = 6$ .
We get $\ln(6^2 - 66) = \ln(-5 \cdot 6)$ , which becomes $\ln(36 - 66) = \ln(-30)$ , or $\ln(-30) = \ln(-30)$ .
Uh oh!
We can't have a negative argument inside a logarithm!
So, $x = 6$ is not a valid solution.

STEP 9

Now, let's move on to the second equation: $8\log(3x + 28) - 5 = 11$ .
First, let's isolate the logarithm by adding $5$ to both sides: $8\log(3x + 28) = 16$ .

STEP 10

Next, divide both sides by $8$ to get $\log(3x + 28) = 2$ .
Remember, when the base of the logarithm isn't written, it's understood to be $10$ .

STEP 11

Now, let's rewrite the equation in exponential form: $10^2 = 3x + 28$ .
This simplifies to $100 = 3x + 28$ .

STEP 12

Subtract $28$ from both sides to get $72 = 3x$ .
Finally, divide both sides by $3$ to find $x = 24$ .

STEP 13

Let's verify this solution.
Plugging $x = 24$ back into the original equation gives us $8\log(3 \cdot 24 + 28) - 5 = 11$ , which simplifies to $8\log(72 + 28) - 5 = 11$ , then $8\log(100) - 5 = 11$ .
Since $\log(100) = 2$ , we have $8 \cdot 2 - 5 = 11$ , or $16 - 5 = 11$ , which is $11 = 11$ .
Perfect!

STEP 14

For the first equation, the solution set is $\{-11\}$ .
For the second equation, the solution set is $\{24\}$ .

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QuestionSolve by using the equivalence property of logarithmic expressions, and write the solution set.
1. $\ln \left(x^{2}-66\right)=\ln (-5 x)$
Solve by using exponential form, and write the solution set.
2. $8 \log (3 x+28)-5=11$

Studdy Solution

STEP 1

STEP 2

1. Solve using equivalence property
2. Solve using exponential form

STEP 3

Alright, let's tackle the first equation: $\ln(x^2 - 66) = \ln(-5x)$ .
Since the logarithms are equal, their arguments must be equal too!
This is the equivalence property!

STEP 4

So, we set $x^2 - 66 = -5x$ .
Now, we've got a quadratic equation.
Let's rearrange it to $x^2 + 5x - 66 = 0$ .

STEP 5

Time to factor!
We're looking for two numbers that multiply to $-66$ and add up to $5$ .
Those numbers are $11$ and $-6$ .
So, we can rewrite the equation as $(x + 11)(x - 6) = 0$ .

STEP 6

This gives us two potential solutions: $x = -11$ and $x = 6$ .
But hold on!
We need to check if these solutions are valid.

STEP 7

Let's check $x = -11$ .
Plugging it back into the original equation, we get $\ln((-11)^2 - 66) = \ln(-5 \cdot (-11))$ , which simplifies to $\ln(121 - 66) = \ln(55)$ , or $\ln(55) = \ln(55)$ .
This works!

STEP 8

Now, let's check $x = 6$ .
We get $\ln(6^2 - 66) = \ln(-5 \cdot 6)$ , which becomes $\ln(36 - 66) = \ln(-30)$ , or $\ln(-30) = \ln(-30)$ .
Uh oh!
We can't have a negative argument inside a logarithm!
So, $x = 6$ is not a valid solution.

STEP 9

Now, let's move on to the second equation: $8\log(3x + 28) - 5 = 11$ .
First, let's isolate the logarithm by adding $5$ to both sides: $8\log(3x + 28) = 16$ .

STEP 10

Next, divide both sides by $8$ to get $\log(3x + 28) = 2$ .
Remember, when the base of the logarithm isn't written, it's understood to be $10$ .

STEP 11

Now, let's rewrite the equation in exponential form: $10^2 = 3x + 28$ .
This simplifies to $100 = 3x + 28$ .

STEP 12

Subtract $28$ from both sides to get $72 = 3x$ .
Finally, divide both sides by $3$ to find $x = 24$ .

STEP 13

STEP 14

For the first equation, the solution set is $\{-11\}$ .
For the second equation, the solution set is $\{24\}$ .

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Studdy Solution

STEP 1

STEP 2

1. Solve using equivalence property2. Solve using exponential form

STEP 3

Alright, let's **tackle the first equation**: ln⁡(x2−66)=ln⁡(−5x)\ln(x^2 - 66) = \ln(-5x)ln(x2−66)=ln(−5x).Since the logarithms are equal, their *arguments* must be equal too!This is the **equivalence property**!

STEP 4

So, we set x2−66=−5xx^2 - 66 = -5xx2−66=−5x.Now, we've got a *quadratic equation*.Let's rearrange it to x2+5x−66=0x^2 + 5x - 66 = 0x2+5x−66=0.

STEP 5

Time to **factor**!We're looking for two numbers that multiply to −66-66−66 and add up to 555.Those numbers are 111111 and −6-6−6.So, we can rewrite the equation as (x+11)(x−6)=0(x + 11)(x - 6) = 0(x+11)(x−6)=0.

STEP 6

This gives us two potential solutions: x=−11x = -11x=−11 and x=6x = 6x=6.But hold on!We need to **check** if these solutions are valid.

STEP 7

STEP 8

STEP 9

Now, let's **move on to the second equation**: 8log⁡(3x+28)−5=118\log(3x + 28) - 5 = 118log(3x+28)−5=11.First, let's **isolate the logarithm** by adding 555 to both sides: 8log⁡(3x+28)=168\log(3x + 28) = 168log(3x+28)=16.

STEP 10

Next, **divide both sides by** 888 to get log⁡(3x+28)=2\log(3x + 28) = 2log(3x+28)=2.Remember, when the base of the logarithm isn't written, it's understood to be 101010.

STEP 11

Now, let's **rewrite the equation in exponential form**: 102=3x+2810^2 = 3x + 28102=3x+28.This simplifies to 100=3x+28100 = 3x + 28100=3x+28.

STEP 12

Subtract 282828 from both sides to get 72=3x72 = 3x72=3x.Finally, **divide both sides by** 333 to find x=24x = 24x=24.

STEP 13

STEP 14

For the first equation, the solution set is {−11}\{-11\}{−11}.For the second equation, the solution set is {24}\{24\}{24}.

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1. Solve using equivalence property
2. Solve using exponential form

Alright, let's tackle the first equation: $\ln(x^2 - 66) = \ln(-5x)$ .
Since the logarithms are equal, their arguments must be equal too!
This is the equivalence property!

So, we set $x^2 - 66 = -5x$ .
Now, we've got a quadratic equation.
Let's rearrange it to $x^2 + 5x - 66 = 0$ .

Time to factor!
We're looking for two numbers that multiply to $-66$ and add up to $5$ .
Those numbers are $11$ and $-6$ .
So, we can rewrite the equation as $(x + 11)(x - 6) = 0$ .

This gives us two potential solutions: $x = -11$ and $x = 6$ .
But hold on!
We need to check if these solutions are valid.

Now, let's move on to the second equation: $8\log(3x + 28) - 5 = 11$ .
First, let's isolate the logarithm by adding $5$ to both sides: $8\log(3x + 28) = 16$ .

Next, divide both sides by $8$ to get $\log(3x + 28) = 2$ .
Remember, when the base of the logarithm isn't written, it's understood to be $10$ .

Now, let's rewrite the equation in exponential form: $10^2 = 3x + 28$ .
This simplifies to $100 = 3x + 28$ .

Subtract $28$ from both sides to get $72 = 3x$ .
Finally, divide both sides by $3$ to find $x = 24$ .

For the first equation, the solution set is $\{-11\}$ .
For the second equation, the solution set is $\{24\}$ .