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Solve for x : (Logarithms)
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\[\log_{4}(3-x)+\log_{0.25}(3+x)= \log_{4}(1-x)+\log_{0.25}(2x+1)\]
x = 0 or 7
solution please!!
actually i m not familiar with equation tool here. so do one thing convert each term to base 2 using log\[\log_{a ^{n}}x = 1/nlog_{a}x \] get all things together leaving zero on one side then its easy to solve
note that 4 = 1/0.25 or log 0.25 = -1 where base is 4 so converting everything to base 4, log (3-x) - log(3+x) = log(1-x) - log(2x+1) => (3-x)/(3+x) = (1-x)(2x+1) just solve for x
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