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What is the solution to the equation = 1 over the square root of 8= 4^(m – 2) ?
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\[\log_6 x + \log_6(x + 1) = 1 \]Now use the property:\[ \log_na + \log_nb = \log_nab\]
hmmm
i messed up by accident ha
\[\frac{1}{\sqrt{8}}=2^{-\frac{3}{2}}\]
\[\left((2^3)^{1 \over 2}\right)^{-1}= (2^2)^{m - 2} \]
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\[2^{-1.5} = 2^{2m - 4} \]Now you can equate the exponents.
swaggg , thanks .
You're welcome.
Dirty answer, isn't it?
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