9. Determine the value of \( x \). \( \begin{array}{lll}\text { a) } 2^{x}=4 & \text { b) } 2^{x}=16 & \text { c) } 10^{x}=1000 \\ \text { d) } 8^{x}=8 & \text { e) } 3^{x}=1 & \text { f) } 4^{x}=64 \\ \text { g) } 5^{x}=625 & \text { h) } 2^{x}=1 & \text { i) } 3^{x}=81\end{array} \)
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Absolutely, let's dive into the solutions for each equation! a) \( 2^{x} = 4 \) Since \( 4 = 2^2 \), we can equate the exponents: \( x = 2 \) b) \( 2^{x} = 16 \) Here, \( 16 = 2^4 \), so: \( x = 4 \) c) \( 10^{x} = 1000 \) Noticing that \( 1000 = 10^3 \): \( x = 3 \) d) \( 8^{x} = 8 \) Recognizing \( 8 = 8^1 \) gives us: \( x = 1 \) e) \( 3^{x} = 1 \) Since any number to the power of 0 equals 1: \( x = 0 \) f) \( 4^{x} = 64 \) Knowing \( 64 = 4^{3} \): \( x = 3 \) g) \( 5^{x} = 625 \) And since \( 625 = 5^4 \): \( x = 4 \) h) \( 2^{x} = 1 \) Again, any number to the power of 0 equals 1: \( x = 0 \) i) \( 3^{x} = 81 \) Since \( 81 = 3^{4} \): \( x = 4 \) Now we have the values of \( x \): - a) \( x = 2 \) - b) \( x = 4 \) - c) \( x = 3 \) - d) \( x = 1 \) - e) \( x = 0 \) - f) \( x = 3 \) - g) \( x = 4 \) - h) \( x = 0 \) - i) \( x = 4 \)
