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if log2=0.301 and log3=0.477, then the number of digits in 6^20 is
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well using log laws 1st deal with the exponent \[\log(x^a) = alog(x) \] apply the above law to the exponent of 20 and the law for multiplication is \[\log(xy) = \log(x) + \log(y) \] where x and y are factors of xy. does this make sense.... ?
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