Solve the following math problems: 1. $5^{-3} = $ 2. $3^x * 9^4 = 3^5$ 3. $log_5 125 - log_3 3 + 2$ 4. $6\sqrt{2} - 2\sqrt{8} + \sqrt{50}$ 5. Find $lg 303 = 0.24771$ find the value... Solve the following math problems: 1. $5^{-3} = $ 2. $3^x * 9^4 = 3^5$ 3. $log_5 125 - log_3 3 + 2$ 4. $6\sqrt{2} - 2\sqrt{8} + \sqrt{50}$ 5. Find $lg 303 = 0.24771$ find the value of $Lg 30$ 6. Rationalize the denominator $\frac{10}{\sqrt{2}}$ 7. If $\frac{a^{-2} * a^8}{a^3} = 8$ Find the value 'a' 8. $\sqrt{5^9 * 2^5} = 20$ 9. $\sqrt[3]{\frac{27}{125}} * \sqrt{\frac{72}{7.50}} * 3^0$ 10. $216^{\frac{1}{3}} = 6^{2} * 6^x$ Read more

Steps to Solve

1. Evaluate $5^{-3}$

A negative exponent means taking the reciprocal and raising to the positive exponent. $5^{-3} = \frac{1}{5^3} = \frac{1}{555} = \frac{1}{125}$

2. Solve for $x$ in $3^x \cdot 9^4 = 3^5$

Rewrite $9$ as $3^2$. $3^x \cdot (3^2)^4 = 3^5$ Simplify the exponents. $3^x \cdot 3^8 = 3^5$ When multiplying exponential terms with the same base, add the exponents. $3^{x+8} = 3^5$ Since the bases are equal, the exponents must be equal. $x + 8 = 5$ Solve for $x$. $x = 5 - 8 = -3$

3. Evaluate $log_5 125 - log_3 3 + 2$

Evaluate each logarithm. $log_5 125 = 3$ because $5^3 = 125$, and $log_3 3 = 1$ because $3^1 = 3$ $3 - 1 + 2$ Simplify. $2 + 2 = 4$

4. Evaluate $6\sqrt{2} - 2\sqrt{8} + \sqrt{50}$

Simplify the radicals. $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$ and $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$ $6\sqrt{2} - 2(2\sqrt{2}) + 5\sqrt{2}$ $6\sqrt{2} - 4\sqrt{2} + 5\sqrt{2}$ Combine like terms. $(6 - 4 + 5)\sqrt{2} = 7\sqrt{2}$

5. Find $lg 30$ given $lg 303 = 0.24771$

The question appears to have a typo. It mentions $lg 303 = 0.24771$ which seems incorrect, as $lg$ refers to base 10 logarithm. Assuming the question meant $log_{10} 3.03=0.4771$, we want to find $log_{10} 30$. $log_{10} 30 = log_{10} (3.03 \cdot \frac{30}{3.03}) = log_{10} 3.03 + log_{10} (\frac{30}{3.03})$ $log_{10} 30 = log_{10}(3.03 \times 10^1) = log_{10}3.03 + log_{10}10$ $log_{10} 30 = log_{10} 3.03 + 1$ $log_{10} 30 \approx 0.4771+1 = 1.4771$ Assuming the given value should have been $log 3.03 \approx 0.4771$ instead of $log 303 = 0.24771$ $log 30 = log (3.03 * \frac{100}{10.1}) = log 3.03 + log (\frac{100}{10.1})$ $log 30 = log (3 * 10) = log 3 + log 10 = log 3+1$ We need to find $log 3$ from the given value and would require more information beyond the given $log 3.03 = 0.4771.$ We will proceed assuming this was a typo

Let's assume the equation should read $log 3 = 0.24771$ then $log 30 = log 3 + log 10 = 0.24771 + 1 = 1.24771$

6. Rationalize the denominator of $\frac{10}{\sqrt{2}}$

Multiply the numerator and denominator by $\sqrt{2}$. $\frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

7. Solve for $a$ if $\frac{a^{-2} \cdot a^8}{a^3} = 8$

Simplify the left side using exponent rules. When multiplying exponential terms with the same base, add the exponents. When dividing, subtract the exponents. $\frac{a^{-2+8}}{a^3} = \frac{a^6}{a^3} = a^{6-3} = a^3$ $a^3 = 8$ Take the cube root of both sides. $a = \sqrt[3]{8} = 2$

8. Simplify $\sqrt{5^9 \cdot 2^5}$

The question states the radical expression equals 20, however it is asking us to simplify it perhaps. $\sqrt{5^9 \cdot 2^5} = \sqrt{5^8 \cdot 5 \cdot 2^4 \cdot 2} = \sqrt{(5^4)^2 \cdot (2^2)^2 \cdot 5 \cdot 2} = 5^4 \cdot 2^2 \cdot \sqrt{10} = 625 \cdot 4 \cdot \sqrt{10} = 2500\sqrt{10}$

9. Evaluate $\sqrt[3]{\frac{27}{125}} \cdot \sqrt{\frac{72}{7.50}} \cdot 3^0$

Simplify each term. $3^0 = 1$. Also, $\sqrt[3]{\frac{27}{125}} = \frac{\sqrt[3]{27}}{\sqrt[3]{125}} = \frac{3}{5}$. Then $\sqrt{\frac{72}{7.5}} = \sqrt{\frac{720}{75}} = \sqrt{\frac{144}{15}} = \sqrt{\frac{144}{15} \cdot \frac{15}{15}} = \sqrt{\frac{144 \cdot 15}{15^2}} = \frac{12\sqrt{15}}{15} = \frac{4\sqrt{15}}{5}$ Multiply: $\frac{3}{5} \cdot \frac{4\sqrt{15}}{5} \cdot 1 = \frac{12\sqrt{15}}{25}$

10. Solve for $x$ in $216^{\frac{1}{3}} = 6^2 \cdot 6^x$

Simplify $216^{\frac{1}{3}}$. Since $6^3 = 216$, $216^{\frac{1}{3}} = 6$. $6 = 6^2 \cdot 6^x$ $6^1 = 6^{2+x}$ Since the bases are equal, the exponents must be equal. $1 = 2 + x$ $x = 1 - 2 = -1$