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g(x) = 64^x - log3(x) , find f(1/3)...

`g(x) = 64^x - log_3(x) `, find `f(1/3)`

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`f(x) = 64^x-log_3(x)`
`:. f(1/3) = 64^(1/3) - log_3(1/3)`
`=> f(1/3) = (4^3)^(1/3) - log_3(3)^-1`
`=> f(1/3) =4 + log_3(3)`
We know, `log_a(a) = 1,`
`:. f(1/3) =4 + 1 = 5`
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