Math / Algebra

QuestionUse properties of logarithms to expand each logarithmic expression as much as possible. Evaluate $\begin{array}{c} \log _{7}(7 x) \\ \log _{7}(7 x)= \end{array}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to rewrite a single logarithm as the sum of simpler logarithms. Watch out! Remember the rules of logarithms, especially the product rule!

STEP 2

1. Expand the logarithm using the product rule.
2. Simplify the expression.

STEP 3

The product rule for logarithms states that the logarithm of a product is the sum of the logarithms of the factors.
Mathematically, this means $\log_b(m \cdot n) = \log_b(m) + \log_b(n)$ .
This rule is super important because it lets us break down complicated logarithms into smaller, easier-to-manage pieces!

STEP 4

In our case, we have $\log_7(7x)$ .
We can think of this as $\log_7(7 \cdot x)$ .
Applying the product rule, we get: $\log_7(7 \cdot x) = \log_7(7) + \log_7(x)$ See how we broke down the logarithm of the product $7x$ into the sum of the logarithms of $7$ and $x$ ?
Awesome!

STEP 5

Remember, $\log_b(a)$ asks the question: "To what power must we raise $b$ to get $a$ ?".

STEP 6

So, $\log_7(7)$ asks: "To what power must we raise $7$ to get $7$ ?".
Well, $7$ raised to the first power is $7$ ( $7^1 = 7$ ), so $\log_7(7) = 1$ !

STEP 7

Now, we can substitute this back into our expanded expression: $\log_7(7) + \log_7(x) = 1 + \log_7(x)$

STEP 8

Our final answer is $1 + \log_7(x)$ .
We've successfully expanded and simplified the logarithm!

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QuestionUse properties of logarithms to expand each logarithmic expression as much as possible. Evaluate $\begin{array}{c} \log _{7}(7 x) \\ \log _{7}(7 x)= \end{array}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to rewrite a single logarithm as the sum of simpler logarithms. Watch out! Remember the rules of logarithms, especially the product rule!

STEP 2

1. Expand the logarithm using the product rule.
2. Simplify the expression.

STEP 3

STEP 4

In our case, we have $\log_7(7x)$ .
We can think of this as $\log_7(7 \cdot x)$ .
Applying the product rule, we get: $\log_7(7 \cdot x) = \log_7(7) + \log_7(x)$ See how we broke down the logarithm of the product $7x$ into the sum of the logarithms of $7$ and $x$ ?
Awesome!

STEP 5

Remember, $\log_b(a)$ asks the question: "To what power must we raise $b$ to get $a$ ?".

STEP 6

So, $\log_7(7)$ asks: "To what power must we raise $7$ to get $7$ ?".
Well, $7$ raised to the first power is $7$ ( $7^1 = 7$ ), so $\log_7(7) = 1$ !

STEP 7

Now, we can substitute this back into our expanded expression: $\log_7(7) + \log_7(x) = 1 + \log_7(x)$

STEP 8

Our final answer is $1 + \log_7(x)$ .
We've successfully expanded and simplified the logarithm!

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QuestionUse properties of logarithms to expand each logarithmic expression as much as possible. Evaluate log⁡7(7x)log⁡7(7x)=\begin{array}{c} \log _{7}(7 x) \\ \log _{7}(7 x)= \end{array}log7​(7x)log7​(7x)=​ □\square□

Studdy Solution

STEP 1

What is this asking? We need to rewrite a single logarithm as the sum of simpler logarithms. Watch out! Remember the rules of logarithms, especially the product rule!

STEP 2

1. Expand the logarithm using the product rule.2. Simplify the expression.

STEP 3

STEP 4

STEP 5

Remember, log⁡b(a)\log_b(a)logb​(a) asks the question: "To what power must we raise bbb to get aaa?".

STEP 6

So, log⁡7(7)\log_7(7)log7​(7) asks: "To what power must we raise 777 to get 777?".Well, 777 raised to the **first** power is 777 (71=77^1 = 771=7), so log⁡7(7)=1\log_7(7) = 1log7​(7)=1!

STEP 7

Now, we can substitute this back into our expanded expression: log⁡7(7)+log⁡7(x)=1+log⁡7(x) \log_7(7) + \log_7(x) = 1 + \log_7(x) log7​(7)+log7​(x)=1+log7​(x)

STEP 8

Our final answer is 1+log⁡7(x)1 + \log_7(x)1+log7​(x).We've successfully expanded and simplified the logarithm!

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QuestionUse properties of logarithms to expand each logarithmic expression as much as possible. Evaluate $\begin{array}{c} \log _{7}(7 x) \\ \log _{7}(7 x)= \end{array}$ $\square$

1. Expand the logarithm using the product rule.
2. Simplify the expression.

Remember, $\log_b(a)$ asks the question: "To what power must we raise $b$ to get $a$ ?".

So, $\log_7(7)$ asks: "To what power must we raise $7$ to get $7$ ?".
Well, $7$ raised to the first power is $7$ ( $7^1 = 7$ ), so $\log_7(7) = 1$ !

Now, we can substitute this back into our expanded expression: $\log_7(7) + \log_7(x) = 1 + \log_7(x)$

Our final answer is $1 + \log_7(x)$ .
We've successfully expanded and simplified the logarithm!