Asked by mathemagiacian
Suppose 81^x = 64. What is 27^{x+1}?
Give me a clear explanation with the answer. Thanks! :)
Give me a clear explanation with the answer. Thanks! :)
Answers
Answered by
Steve
clearly, x = log<sub>81</sub>64 = 1/4 log<sub>3</sub>64
27^{x+1} = 27*3^3x
= 27*3^(3(1/4 log<sub>3</sub>64))
= 27*3^(3/4 log<sub>3</sub>64)
= 27*3^((log<sub>3</sub>64) * 3/4)
= 27*64^(3/4)
= 27*2^(9/2)
= 27*16√2
= 432√2
27^{x+1} = 27*3^3x
= 27*3^(3(1/4 log<sub>3</sub>64))
= 27*3^(3/4 log<sub>3</sub>64)
= 27*3^((log<sub>3</sub>64) * 3/4)
= 27*64^(3/4)
= 27*2^(9/2)
= 27*16√2
= 432√2
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