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what is 4^-2=1 over 16 in logarithmic form?
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\(\large 4^{-2} = \dfrac{1}{16} \) \(\large a^x =b \implies \log_a b=x\)
can you explain how i would work this out
sure, take log on both sides of the equation, what do u get ?
so take the log of both sides
yup
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well the log of 4^-2 is -1.204
we need not evaluate it. we need the log form, so keep it as log 4^-2 now use the fact that log a^b = b log a then log 4^-2 = ... ?
and the log of 1 over 16 is the same thing
ofcourse, 4^-2 = 1/4^2 = 1/16 so their log will be same
so my answer is the log of 4^-2=1/16 right
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the answer should be in log form let me write out the steps
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