EXPS 1 | Lesson 4 | Practice Solutions

\(2^x=32\), then \(2^x=2^5 ; x=5\)
\(2^x=4^2\), then \(2^x=(2^2)^2\), then \(2^x=2^4 ; x=4\)
\(4^{x-1}=16^3\), then \(4^{x-1}=(4^2)^3\), then \(4^{x-1}=4^6\), then \(x-1=6 ; x=7\)
\(5^{\frac x2}=125\), then \(5^{\frac x2}=5^3\), then \(\dfrac x2= 3 ; x =6\)
\(3^{2-x}=27\), then \(3^{2-x}=3^3\), then \(2-x=3 ; x=-1\)
\(2^{x+2}=8^x\), then \(2^{x+2}=(2^3)^x\), then \(2^{x+2}=2^{3x}\), then \(x+2=3x ; x=1\)
\(2^{2x-1}=8^3\), then \(2^{2x-1}=(2^3)^3\), then \(2x-1=9 ; x=5\)
\(2^{3x-1}=4^{x+3}\), then \(2^{3x-1}=(2^2)^{x+3}\), then \(2^{3x-1}=2^{2x+6}\), then \(3x-1=2x+6 ; x=7\)

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