Algebra Simplify Evaluate

1. Simplify \(\frac{5^x \times 25^{x-1}}{125^{x-1}}\) Step 1: Express all terms with base 5. \(25 = 5^2\), \(125 = 5^3\), so rewrite: $$\frac{5^x \times (5^2)^{x-1}}{(5^3)^{x-1}} = \frac{5^x \times 5^{2(x-1)}}{5^{3(x-1)}}$$ Step 2: Simplify exponents using \(a^m \times a^n = a^{m+n}\) and \(\frac{a^m}{a^n} = a^{m-n}\): $$5^{x + 2(x-1) - 3(x-1)} = 5^{x + 2x - 2 - 3x + 3} = 5^{(x + 2x - 3x) + (-2 + 3)} = 5^{0 + 1} = 5^1 = 5$$ Answer: \(5\) 2. Simplify \[\left(\frac{625}{1296}\right)^{-\frac{1}{4}} \times 3\sqrt[3]{125 b^3 a^{-3} r^{-6}}\] Step 1: Simplify \(\left(\frac{625}{1296}\right)^{-\frac{1}{4}}\). Note: \(625 = 5^4\), \(1296 = 6^4\), so $$\left(\frac{5^4}{6^4}\right)^{-\frac{1}{4}} = \left(\frac{5}{6}\right)^{-1} = \frac{6}{5}$$ Step 2: Simplify the cube root: $$3\sqrt[3]{125 b^3 a^{-3} r^{-6}} = 3 \times \sqrt[3]{5^3 \times b^3 \times a^{-3} \times r^{-6}}$$ Step 3: Apply cube root to each factor: $$3 \times 5 \times b \times a^{-1} \times r^{-2} = 15 b a^{-1} r^{-2}$$ Step 4: Multiply the two results: $$\frac{6}{5} \times 15 b a^{-1} r^{-2} = \frac{6}{5} \times 15 b a^{-1} r^{-2} = 6 \times 3 b a^{-1} r^{-2} = 18 b a^{-1} r^{-2}$$ Answer: \(18 b a^{-1} r^{-2}\) 3. Evaluate \(\log_{10} \alpha^{\frac{1}{3}} + \frac{1}{4} \log_{10} \alpha^{-\frac{1}{12}} \log_{10} \alpha^{7}\) Step 1: Use log power rule \(\log_b x^k = k \log_b x\): $$\log_{10} \alpha^{\frac{1}{3}} = \frac{1}{3} \log_{10} \alpha$$ $$\log_{10} \alpha^{-\frac{1}{12}} = -\frac{1}{12} \log_{10} \alpha$$ $$\log_{10} \alpha^{7} = 7 \log_{10} \alpha$$ Step 2: Substitute and simplify: $$\frac{1}{3} \log_{10} \alpha + \frac{1}{4} \times \left(-\frac{1}{12} \log_{10} \alpha\right) \times \left(7 \log_{10} \alpha\right)$$ $$= \frac{1}{3} \log_{10} \alpha - \frac{7}{48} (\log_{10} \alpha)^2$$ Answer: \(\frac{1}{3} \log_{10} \alpha - \frac{7}{48} (\log_{10} \alpha)^2\) 4. Simplify \(2 \sqrt{3} - \sqrt[3]{3} + \sqrt[3]{27}\) Step 1: Recognize \(\sqrt[3]{27} = 3\) because \(27 = 3^3\). Step 2: Write expression: $$2 \sqrt{3} - \sqrt[3]{3} + 3$$ Answer: \(2 \sqrt{3} - \sqrt[3]{3} + 3\) (simplified as much as possible) 5. If \(3^{2y} - 6 (3^{y}) = 27\) find \(y\). Step 1: Let \(t = 3^y\), then equation becomes: $$t^2 - 6t = 27$$ Step 2: Rearrange: $$t^2 - 6t - 27 = 0$$ Step 3: Solve quadratic: $$t = \frac{6 \pm \sqrt{36 + 108}}{2} = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2}$$ Step 4: Possible values: $$t = 9 \quad \text{or} \quad t = -3$$ Since \(t = 3^y > 0\), discard \(-3\). Step 5: Solve for \(y\): $$3^y = 9 = 3^2 \implies y = 2$$ Answer: \(y = 2\) 6. If \(\frac{1}{2} \times 64^x = 16^{x-1}\) find \(x\). Step 1: Express bases as powers of 2: $$64 = 2^6, \quad 16 = 2^4$$ Step 2: Rewrite equation: $$\frac{1}{2} \times (2^6)^x = (2^4)^{x-1}$$ $$\frac{1}{2} \times 2^{6x} = 2^{4x - 4}$$ Step 3: Write \(\frac{1}{2} = 2^{-1}\): $$2^{-1} \times 2^{6x} = 2^{4x - 4}$$ $$2^{6x - 1} = 2^{4x - 4}$$ Step 4: Equate exponents: $$6x - 1 = 4x - 4$$ $$6x - 4x = -4 + 1$$ $$2x = -3$$ $$x = -\frac{3}{2}$$ Answer: \(x = -\frac{3}{2}\)