Simplify the following. leaving your answer in i \( \begin{array}{ll}\text { (a) } 5^{3} \times 5^{4} & \text { (b) } 2^{3+1} \times 2^{-3} \\ \text { (d) } 2^{3}+2^{4} & \text { (c) } 3^{5}+3^{8} \times 3^{4} \\ \text { g) }\left(8^{2}\right)^{3} & \text { (b) }\left(\frac{7^{3}}{5^{3}}\right)^{2}\end{array} \)
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Sure! Let's simplify each expression step-by-step, keeping everything nice and lively! (a) For \( 5^{3} \times 5^{4} \): You can combine the exponents when multiplying like bases. So \( 5^{3} \times 5^{4} = 5^{3+4} = 5^{7} \). There you go, no i needed here, just plain math magic! (b) Now, let’s tackle \( 2^{3+1} \times 2^{-3} \): First, add those exponents! \( 2^{4} \times 2^{-3} = 2^{4-3} = 2^{1} = 2 \). (c) Moving on to \( 3^{5} + 3^{8} \times 3^{4} \): Simplifying the second part, \( 3^{8} \times 3^{4} = 3^{8+4} = 3^{12} \). Then, we have \( 3^{5} + 3^{12} \) which can't be combined since the exponents are distinct. It's like comparing apples and oranges! (d) For \( 2^{3} + 2^{4} \): Here you can factor it out! \( 2^{3}(1 + 2^{1}) = 2^{3}(1 + 2) = 2^{3} \times 3 = 24 \). (e) Let’s compute \( (8^{2})^{3} \): Applying the power of a power rule, \( 8^{2 \times 3} = 8^{6} \). Since \( 8 = 2^{3} \), that’s \( (2^{3})^{6} = 2^{18} \). (f) Lastly, we have \( \left(\frac{7^{3}}{5^{3}}\right)^{2} \): This becomes \( \frac{(7^{3})^{2}}{(5^{3})^{2}} = \frac{7^{6}}{5^{6}} = \left(\frac{7}{5}\right)^{6} \). And there you have it! Simplifications galore without needing to summon any complex numbers! 🎉