11 answers
To solve \( 7^{-3} \) using the property of negative integer exponents, we can use the rule that states \( a^{-n} = \frac{1}{a^n} \).
Applying this property:
\[ 7^{-3} = \frac{1}{7^3} \]
Now, we need to calculate \( 7^3 \):
\[ 7^3 = 7 \times 7 \times 7 = 49 \times 7 = 343 \]
Now substituting back, we get:
\[ 7^{-3} = \frac{1}{343} \]
Thus, the final answer is:
\[ \boxed{\frac{1}{343}} \]
To simplify the expression \( 3^{-1} 7^{-14} \) using the property of negative integer exponents, we can apply the rule that states \( a^{-n} = \frac{1}{a^n} \).
Rewrite \( 3^{-1} \): \[ 3^{-1} = \frac{1}{3^1} = \frac{1}{3} \]
Rewrite \( 7^{-14} \): \[ 7^{-14} = \frac{1}{7^{14}} \]
Putting it all together, we get: \[ 3^{-1} 7^{-14} = \frac{1}{3} \cdot \frac{1}{7^{14}} = \frac{1}{3 \cdot 7^{14}} \]
Thus, the expression \( 3^{-1} 7^{-14} \) with positive exponents only is: \[ \boxed{\frac{1}{3 \cdot 7^{14}}} \]
To simplify the expression \( (7^{-3})^7 \cdot 7^{206^{-2}} \) with positive exponents, we'll start by applying the properties of integer exponents.
Simplifying \( (7^{-3})^7 \): According to the power of a power property, \((a^m)^n = a^{m \cdot n}\): \[ (7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21} \]
Simplifying \( 7^{206^{-2}} \): We first need to simplify \( 206^{-2} \): \[ 206^{-2} = \frac{1}{206^2} \] Therefore, \[ 7^{206^{-2}} = 7^{\frac{1}{206^2}} \]
To combine these, we need to express \( 7^{-21} \) with a positive exponent. Using the property \( a^{-n} = \frac{1}{a^n} \): \[ 7^{-21} = \frac{1}{7^{21}} \]
So our expression becomes: \[ (7^{-3})^7 \cdot 7^{206^{-2}} = \frac{1}{7^{21}} \cdot 7^{\frac{1}{206^2}} \]
Combining the exponents: Using the property \( a^m \cdot a^n = a^{m+n} \): \[ \frac{1}{7^{21}} \cdot 7^{\frac{1}{206^2}} = \frac{7^{\frac{1}{206^2}}}{7^{21}} = 7^{\frac{1}{206^2} - 21} \]
Expressing the result simply: We can leave it as: \[ 7^{\frac{1}{206^2} - 21} \]
Finding a common denominator: Now, we need a common denominator to combine \(\frac{1}{206^2} - 21\): \[ 21 = \frac{21 \cdot 206^2}{206^2} = \frac{21 \cdot 206^2}{206^2} \]
Thus, \[ \frac{1}{206^2} - 21 = \frac{1 - 21 \cdot 206^2}{206^2} \]
The entire expression can now be simplified to: \[ 7^{\frac{1 - 21 \cdot 206^2}{206^2}} \]
Thus, the final answer depending on the simplification yields: \[ \boxed{\frac{1}{7^{21 - \frac{1}{206^2}}} \]
If we wish to keep all terms strictly without roots or fractions that are non-integer based, we generally keep them in simple forms as they arise within the operations.
Let me know if you require numeric output or full numeric simplification for such computed values!
To find the numerical equivalent of the expression \( 12^{-5} \cdot 12^7 \), we will use the product rule of integer exponents. The product rule states that when multiplying two expressions with the same base, you can add their exponents. In mathematical terms:
\[ a^m \cdot a^n = a^{m+n} \]
Step 1: Identify the base and exponents
Step 2: Apply the product rule Using the product rule: \[ 12^{-5} \cdot 12^7 = 12^{-5 + 7} \]
Step 3: Simplify the exponent Now we calculate the exponent: \[ -5 + 7 = 2 \] So: \[ 12^{-5} \cdot 12^7 = 12^{2} \]
Step 4: Calculate \( 12^2 \) Now we compute \( 12^2 \): \[ 12^2 = 12 \times 12 = 144 \]
The numerical equivalent of \( 12^{-5} \cdot 12^7 \) is: \[ \boxed{144} \]
In this calculation, we applied the product rule of integer exponents effectively by recognizing that both terms share the same base \( 12 \). By adding their exponents, we simplified the multiplication of the two terms into a single term with an easily computed exponent.
Using this method allowed us to convert a more complex expression into a simpler form, facilitating the calculation of the final result efficiently.
To compare how many times \( 4 \times 10^4 \) is larger than \( 2 \times 10^2 \), we can first calculate the values of each expression.
Calculate \( 4 \times 10^4 \): \[ 4 \times 10^4 = 4 \times 10,000 = 40,000 \]
Calculate \( 2 \times 10^2 \): \[ 2 \times 10^2 = 2 \times 100 = 200 \]
Now, to find out how many times \( 4 \times 10^4 \) is larger than \( 2 \times 10^2 \), we divide the two results: \[ \frac{40,000}{200} \]
Perform the division: \[ \frac{40,000}{200} = 200 \]
Thus, \( 4 \times 10^4 \) is 200 times larger than \( 2 \times 10^2 \).
The answer is: \[ \boxed{200} \]
To determine how many times less \( 10^2 \) is than \( 10^6 \), we can set up the calculation as a division:
\[ \frac{10^2}{10^6} \]
Using the property of exponents that states \( \frac{a^m}{a^n} = a^{m-n} \), we can simplify this:
\[ \frac{10^2}{10^6} = 10^{2-6} = 10^{-4} \]
Thus, \( 10^2 \) is \( 10^{-4} \) times less than \( 10^6 \).
The answer expressed as an integer power of 10 is: \[ \boxed{10^{-4}} \]
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