Example Question - exponents
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Understanding Exponential Expressions
<p>To simplify the expression, we use the property of exponents that states:</p> <p>When dividing like bases, you subtract the exponents:</p> <p>\(\frac{x^a}{x^b} = x^{a-b}\)</p> <p>Thus, the equation holds:</p> <p>\(\frac{x^a}{x^b} = x^{a-b}\)</p>
Simplifying Algebraic Expressions
<p>Para simplificar la expresión dada:</p> <p>Comenzamos con:</p> <p> \(\frac{x^2 \cdot (y^5)^3 \cdot \left( \frac{1}{z} \right)^2}{\frac{x^3}{z^5} \cdot \left( \frac{y^2}{z} \right)^{-4}}\)</p> <p>Reescribiendo y simplificando paso a paso:</p> <p>Numerador:</p> <p> \(x^2 \cdot y^{15} \cdot \frac{1}{z^2}\)</p> <p>Denominador:</p> <p> \(\frac{x^3}{z^5} \cdot \frac{z^4}{y^8}\)</p> <p>Combinando:</p> <p> \(\frac{x^2 \cdot y^{15} \cdot z^4}{x^3 \cdot y^8 \cdot z^2 \cdot z^5}\)</p> <p>Reduciendo términos:</p> <p> \(= \frac{y^{15 - 8}}{z^{2 + 5 - 4} \cdot x^{3 - 2}}\)</p> <p>Resultado final:</p> <p> \(\frac{y^7}{z^5 \cdot x}\)</p>
Simplifying an Expression Involving Exponents
<p>Given the expression:</p> <p>\(\frac{10^{-a}}{7^5 \times 10^7 \times 7^{-7}}\)</p> <p>First, simplify the denominator:</p> <p>\(7^5 \times 7^{-7} = 7^{5 - 7} = 7^{-2}\)</p> <p>Now, rewrite the entire expression:</p> <p>\(\frac{10^{-a}}{7^{-2} \times 10^7}\)</p> <p>This can be rewritten as:</p> <p> \(\frac{10^{-a}}{10^7} \times 7^{2}\)</p> <p>Now, simplify the powers of 10:</p> <p> \(10^{-a - 7} \times 7^2\)</p> <p>Thus, the final simplified expression is:</p> <p>\(7^2 \times 10^{-(a + 7)}\)</p> ```
Finding the Value of an Exponent
<p>Given the equation:</p> <p>$$9^{2m - 5} \times 9^3 = 9^{m + 1}$$</p> <p>Combine the left-hand side using the property of exponents:</p> <p>$$9^{(2m - 5) + 3} = 9^{m + 1}$$</p> <p>This simplifies to:</p> <p>$$9^{2m - 2} = 9^{m + 1}$$</p> <p>Since the bases are the same, set the exponents equal:</p> <p>$$2m - 2 = m + 1$$</p> <p>Isolate \( m \):</p> <p>$$2m - m = 1 + 2$$</p> <p>$$m = 3$$</p>
Expressing a Number in Exponential Form
<p>Given \( 65^{\frac{4}{5}} = b \cdot \sqrt{a^c} \), we need to express the given equation in terms of \( a \), \( b \), and \( c \).</p> <p>From the equation:</p> <p>Let \( a = 65 \), \( b = 4 \), and \( c = 5 \).</p> <p>Therefore, the values are:</p> <p> \( a = 65 \), \( b = 4 \), \( c = 5 \).</p>
Simplifying a Fraction with Exponents
<p>Given the expression:</p> <p>\[\frac{10^{-8}}{7^5 \times 10^3 \times 7^{-7}}\]</p> <p>We can rewrite the denominator:</p> <p>\(7^5 \times 10^3 \times 7^{-7} = 10^3 \times 7^{5 - 7} = 10^3 \times 7^{-2}\)</p> <p>Next, we place this back into the expression:</p> <p>\[\frac{10^{-8}}{10^3 \times 7^{-2}} = \frac{10^{-8}}{10^3} \times 7^2\]</p> <p>Simplifying \(\frac{10^{-8}}{10^3}\):</p> <p>\(10^{-8 - 3} = 10^{-11}\)</p> <p>The expression becomes:</p> <p>\(10^{-11} \times 7^2\)</p> <p>Thus, the final simplified expression is:</p> <p>\[7^2 \times 10^{-11}\]</p> <p>Which can be written as:</p> <p>\(49 \times 10^{-11}\)</p>
Understanding Powers and Exponents
<p>To simplify the expression, we recognize the pattern in the powers of -16.</p> <p>The expression can be rewritten as:</p> <p>(-16)<sup>1</sup> × (-16)<sup>2</sup> × (-16)<sup>3</sup> × (-16)<sup>4</sup> × (-16)<sup>5</sup></p> <p>This is equivalent to:</p> <p>(-16)<sup>1+2+3+4+5</sup></p> <p>Calculating the exponent:</p> <p>1 + 2 + 3 + 4 + 5 = 15</p> <p>So, the simplified expression is:</p> <p>(-16)<sup>15</sup></p>
Exercises on Exponents
<p>Para resolver los ejercicios de potencias, utilizaremos las propiedades de las potencias. </p> <p>Para el ejercicio \(2^{-3} \cdot 2^{5} \cdot 2^{-10}\): </p> <p>Aplicamos la propiedad de producto de potencias con la misma base sumando los exponentes: </p> <p>\(2^{-3 + 5 - 10} = 2^{-8}\) </p> <p>Por lo tanto, la respuesta simplificada es \(2^{-8}\).</p> <p>Para el ejercicio \((2^0)^{3} \cdot 5^{0}\): </p> <p>Recordamos que cualquier número elevado a la potencia de cero es igual a uno: </p> <p>\((2^0)^{3} = 1^3\) y \(5^{0} = 1\)</p> <p>Por lo tanto, la respuesta simplificada es \(1^3 \cdot 1 = 1\).</p> <p>Para el ejercicio \(8^{-1} \cdot 8^2\): </p> <p>Aplicamos la propiedad de producto de potencias con la misma base sumando los exponentes: </p> <p>\(8^{-1 + 2} = 8^1\)</p> <p>Por lo tanto, la respuesta simplificada es \(8^1\) o simplemente \(8\).</p> <p>Para el ejercicio \(5^{x+1}\): </p> <p>Esta expresión ya está simplificada y no se puede simplificar más sin conocer el valor de x.</p> <p>Para el ejercicio \(x^{-1} \cdot x\): </p> <p>Aplicamos la propiedad de producto de potencias con la misma base sumando los exponentes:</p> <p>\(x^{-1 + 1} = x^0\)</p> <p>Y sabemos que cualquier número elevado a la potencia de cero es igual a uno: \(x^0 = 1\).</p>
Simplifying a Fraction with Exponents
<p>Wir haben den Ausdruck \(\left(\frac{2}{3}\right)^5 \div \left(\frac{7}{10}\right)^2\) und möchten diesen vereinfachen.</p> <p>Schritt 1: Schreibe den Ausdruck mit negativen Exponenten um, um die Division durch Multiplikation zu ersetzen:</p> <p>\(\left(\frac{2}{3}\right)^5 \cdot \left(\frac{10}{7}\right)^2\).</p> <p>Schritt 2: Wende die Potenzregeln beim Multiplizieren von Potenzen mit gleicher Basis an:</p> <p>\(\frac{2^5}{3^5} \cdot \frac{10^2}{7^2}\).</p> <p>Schritt 3: Multipliziere die Zähler und die Nenner miteinander:</p> <p>\(\frac{2^5 \cdot 10^2}{3^5 \cdot 7^2}\).</p> <p>Schritt 4: Führe die Potenzrechnung durch:</p> <p>\(\frac{32 \cdot 100}{243 \cdot 49}\).</p> <p>Schritt 5: Multipliziere die Zahlen aus:</p> <p>\(\frac{3200}{11907}\).</p> <p>Dies ist die vereinfachte Form des gegebenen Ausdrucks.</p>
Simplifying Exponents and Bases
<p>\[\frac{5^4 \cdot 9^5}{15^3 \cdot 27^2}\]</p> <p>Bước 1: Định dạng lại biểu thức theo cơ số chung.</p> <p>\[= \frac{5^4 \cdot (3^2)^5}{(3 \cdot 5)^3 \cdot (3^3)^2}\]</p> <p>\[= \frac{5^4 \cdot 3^{10}}{3^3 \cdot 5^3 \cdot 3^6}\]</p> <p>Bước 2: Rút gọn biểu thức sử dụng luật số mũ.</p> <p>\[= \frac{5^4}{5^3} \cdot \frac{3^{10}}{3^9}\]</p> <p>\[= 5^{4-3} \cdot 3^{10-9}\]</p> <p>Bước 3: Tính toán kết quả cuối cùng.</p> <p>\[= 5^1 \cdot 3^1\]</p> <p>\[= 5 \cdot 3\]</p> <p>\[= 15\]</p>
Exponents and Their Multiplicative Representation
<p>Langkah pertama adalah memahami notasi eksponensial:</p> <p>\(a^n\) berarti bahwa \(a\) dikalikan dengan dirinya sendiri sebanyak \(n\) kali.</p> <p>Maka solusi lengkapnya:</p> <p>\(a^n = a \times a \times a \times ... \times a\) (sebanyak \(n\) faktor)</p>
Algebraic Evaluation of an Expression with Exponents
Given the equation \(2^x = 3^y = 6^z\), we want to find the value of \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\). <p>\(6^z = (2 \cdot 3)^z = 2^z \cdot 3^z\)</p> <p>Since \(2^x = 3^y = 6^z\), we can say \(2^x = 2^z \cdot 3^z\).</p> <p>Therefore, \(x = z \cdot (\log_2{2} + \log_2{3}) = z + z \cdot \log_2{3}\).</p> <p>Similarly, \(3^y = 2^z \cdot 3^z\) implies \(y = z \cdot (\log_3{2} + \log_3{3}) = z \cdot \log_3{2} + z\).</p> <p>We find \(\log_2{3}\) and \(\log_3{2}\) by changing the base:</p> <p>\(\log_2{3} = \frac{1}{\log_3{2}}\)</p> <p>\(x = z + z \cdot \log_2{3} = z + \frac{z}{\log_3{2}}\)</p> <p>\(y = z \cdot \log_3{2} + z\)</p> <p>We then calculate the sum:</p> <p>\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{z + \frac{z}{\log_3{2}}} + \frac{1}{z \cdot \log_3{2} + z} + \frac{1}{z}\)</p> <p>By finding a common denominator, we have:</p> <p>\(\frac{\log_3{2}(\log_3{2} + 1) + 1 + \log_3{2}(\log_3{2} + 1)}{z(\log_3{2} + 1)}\)</p> <p>This simplifies to:</p> <p>\(\frac{2\log_3{2}(\log_3{2} + 1) + 1}{z(\log_3{2} + 1)}\)</p> <p>Since \(\log_3{2}(\log_3{2} + 1) = 1\), we get:</p> <p>\(\frac{2 \cdot 1 + 1}{z(\log_3{2} + 1)} = \frac{3}{z(\log_3{2} + 1)}\)</p> <p>Now, \(y = z \cdot \log_3{2} + z\) gives us \(y = z(\log_3{2} + 1)\), hence substituting \(y\) into the denominator yields:</p> <p>\(\frac{3}{y}\)</p> <p>Since \(2^x=3^y\), it follows that \(x=y\), thus \(\frac{3}{y} = \frac{3}{x}\).</p> <p>Finally, \(2^x=3^y\) implies \(1 = \frac{3}{x}\), so the value of the sum \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) is 3.</p>
Simplification of Algebraic Expression with Exponents
<p>\(\frac{x^7 x^{-n}}{x^5 x^{-2n}} = \frac{x^{7-n}}{x^{5-2n}}\)</p> <p>\(= x^{(7-n)-(5-2n)}\)</p> <p>\(= x^{7-n-5+2n}\)</p> <p>\(= x^{2+n}\)</p>
Simplifying a Fraction with Exponents
Given the expression \(\frac{4a^6b^5c^{-2}}{(2a^{-4}b^6c)^3}\), simplify as follows: \[ \left(\frac{4a^6b^5c^{-2}}{(2a^{-4}b^6c)^3}\right) = \frac{4a^6b^5c^{-2}}{8a^{-12}b^{18}c^3} \] Apply the exponent rule \(a^m \cdot a^n = a^{m+n}\) and simplify: \[ = 4a^{6 - (-12)}b^{5 - 18}c^{-2 - 3} \] \[ = 4a^{18}b^{-13}c^{-5} \] \[ = \frac{4a^{18}}{b^{13}c^{5}} \] Thus, the final simplified form is \(\frac{4a^{18}}{b^{13}c^{5}}\).
Finding Closest Integer to Cube Root
The expression in the question is the cube root of √2, which can be rewritten using exponents as (2^(1/2))^(1/3). When we multiply the exponents, we get 2^(1/6). To find out which integer this is closest to, we can look for small integers and their sixth powers to compare. The sixth power of 1 is 1 (since 1^6 = 1), and the sixth power of 2 is 64 (since 2^6 = 64). The number 2^(1/6) will be somewhere between 1 and 2, but closer to 1 since 2 is much larger when raised to the sixth power. Therefore, the integer closest to the cube root of √2 is 1.
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