Asked by Angel
Suppose 2^x = 10.
Compute 8^x
i get 8 ^((ln 10) / (ln 2) ) but this was incorrect? i do think it was right, ideas or something im missing?
Compute 8^x
i get 8 ^((ln 10) / (ln 2) ) but this was incorrect? i do think it was right, ideas or something im missing?
Answers
Answered by
R_scott
8 = 2^3
(2^3)^x = (2^x)^3 = 10^3
(2^3)^x = (2^x)^3 = 10^3
Answered by
oobleck
recall how to change base of logs.
8 ^((ln 10) / (ln 2) ) = 8^log<sub><sub>2</sub></sub>10
= (2^3)^<sub><sub>2</sub></sub>10
= 2^(<sub><sub>2</sub></sub>10)^3
But, 2^<sub><sub>2</sub></sub>10 = 10
so, it's still 10^3
8 ^((ln 10) / (ln 2) ) = 8^log<sub><sub>2</sub></sub>10
= (2^3)^<sub><sub>2</sub></sub>10
= 2^(<sub><sub>2</sub></sub>10)^3
But, 2^<sub><sub>2</sub></sub>10 = 10
so, it's still 10^3
Answered by
Henry2
2^x = 10.
x*Log2 = Log10,
X = Log10/Log2 = 3.32192809.
8^x = 8^(3.32192809) = 1000.
x*Log2 = Log10,
X = Log10/Log2 = 3.32192809.
8^x = 8^(3.32192809) = 1000.
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