Math / Algebra

QuestionUse the elementary properties of logarithms to solve the following equation. Write your answer as a fraction reduced to lowest terms. $\log _{2}(7 x)=5$

Studdy Solution

STEP 1

What is this asking? This problem is asking us to find the value of $x$ that makes the equation $\log_2(7x) = 5$ true. Watch out! Remember the fundamental relationship between logarithms and exponents!
Don't mix up the base and the argument.

STEP 2

1. Rewrite the logarithmic equation in exponential form.
2. Isolate $x$ .

STEP 3

Alright, let's kick things off by remembering what a logarithm actually means.
The equation $\log_2(7x) = 5$ is asking, "2 raised to what power equals $7x$ ?".
The answer, of course, is 5!

STEP 4

So, we can rewrite our logarithmic equation in exponential form: $2^5 = 7x$ This is much easier to work with!

STEP 5

Now, we want to get $x$ all by itself.
Let's evaluate $2^5$ : $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$ So, our equation becomes: $32 = 7x$

STEP 6

To isolate $x$ , we need to divide both sides of the equation by 7.
Remember, we're dividing by 7 so that the 7 on the right-hand side becomes 1 when multiplied by its inverse $\frac{1}{7}$ . $\frac{32}{7} = \frac{7x}{7}$

STEP 7

This simplifies to: $\frac{32}{7} = 1 \cdot x$ $\frac{32}{7} = x$ And there we have it!

STEP 8

Our final answer is $x = \frac{32}{7}$ .
Boom!

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QuestionUse the elementary properties of logarithms to solve the following equation. Write your answer as a fraction reduced to lowest terms. $\log _{2}(7 x)=5$

Studdy Solution

STEP 1

What is this asking? This problem is asking us to find the value of $x$ that makes the equation $\log_2(7x) = 5$ true. Watch out! Remember the fundamental relationship between logarithms and exponents!
Don't mix up the base and the argument.

STEP 2

1. Rewrite the logarithmic equation in exponential form.
2. Isolate $x$ .

STEP 3

Alright, let's kick things off by remembering what a logarithm actually means.
The equation $\log_2(7x) = 5$ is asking, "2 raised to what power equals $7x$ ?".
The answer, of course, is 5!

STEP 4

So, we can rewrite our logarithmic equation in exponential form: $2^5 = 7x$ This is much easier to work with!

STEP 5

Now, we want to get $x$ all by itself.
Let's evaluate $2^5$ : $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$ So, our equation becomes: $32 = 7x$

STEP 6

To isolate $x$ , we need to divide both sides of the equation by 7.
Remember, we're dividing by 7 so that the 7 on the right-hand side becomes 1 when multiplied by its inverse $\frac{1}{7}$ . $\frac{32}{7} = \frac{7x}{7}$

STEP 7

This simplifies to: $\frac{32}{7} = 1 \cdot x$ $\frac{32}{7} = x$ And there we have it!

STEP 8

Our final answer is $x = \frac{32}{7}$ .
Boom!

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QuestionUse the elementary properties of logarithms to solve the following equation. Write your answer as a fraction reduced to lowest terms. log⁡2(7x)=5\log _{2}(7 x)=5log2​(7x)=5

Studdy Solution

STEP 1

What is this asking? This problem is asking us to find the value of xxx that makes the equation log⁡2(7x)=5\log_2(7x) = 5log2​(7x)=5 true. Watch out! Remember the fundamental relationship between logarithms and exponents!Don't mix up the base and the argument.

STEP 2

1. Rewrite the logarithmic equation in exponential form.2. Isolate xxx.

STEP 3

Alright, let's **kick things off** by remembering what a logarithm *actually means*.The equation log⁡2(7x)=5\log_2(7x) = 5log2​(7x)=5 is asking, "2 raised to what power equals 7x7x7x?".The answer, of course, is **5**!

STEP 4

So, we can **rewrite** our logarithmic equation in *exponential form*: 25=7x2^5 = 7x25=7x This is much easier to work with!

STEP 5

Now, we want to get xxx all by itself.Let's **evaluate** 252^525: 25=2⋅2⋅2⋅2⋅2=322^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 3225=2⋅2⋅2⋅2⋅2=32 So, our equation becomes: 32=7x32 = 7x32=7x

STEP 6

To **isolate** xxx, we need to divide *both sides* of the equation by **7**.Remember, we're dividing by 7 so that the 7 on the right-hand side becomes 1 when multiplied by its inverse 17\frac{1}{7}71​. 327=7x7\frac{32}{7} = \frac{7x}{7}732​=77x​

STEP 7

This **simplifies** to: 327=1⋅x\frac{32}{7} = 1 \cdot x732​=1⋅x 327=x\frac{32}{7} = x732​=xAnd there we have it!

STEP 8

Our **final answer** is x=327x = \frac{32}{7}x=732​.Boom!

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionUse the elementary properties of logarithms to solve the following equation. Write your answer as a fraction reduced to lowest terms. $\log _{2}(7 x)=5$

What is this asking? This problem is asking us to find the value of $x$ that makes the equation $\log_2(7x) = 5$ true. Watch out! Remember the fundamental relationship between logarithms and exponents!
Don't mix up the base and the argument.

1. Rewrite the logarithmic equation in exponential form.
2. Isolate $x$ .

Alright, let's kick things off by remembering what a logarithm actually means.
The equation $\log_2(7x) = 5$ is asking, "2 raised to what power equals $7x$ ?".
The answer, of course, is 5!

So, we can rewrite our logarithmic equation in exponential form: $2^5 = 7x$ This is much easier to work with!

Now, we want to get $x$ all by itself.
Let's evaluate $2^5$ : $2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$ So, our equation becomes: $32 = 7x$

To isolate $x$ , we need to divide both sides of the equation by 7.
Remember, we're dividing by 7 so that the 7 on the right-hand side becomes 1 when multiplied by its inverse $\frac{1}{7}$ . $\frac{32}{7} = \frac{7x}{7}$

This simplifies to: $\frac{32}{7} = 1 \cdot x$ $\frac{32}{7} = x$ And there we have it!

Our final answer is $x = \frac{32}{7}$ .
Boom!