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Mathematics 18 Online
OpenStudy (anonymous):

Solve for x : (Logarithms)

OpenStudy (anonymous):

\[\log_{4}(3-x)+\log_{0.25}(3+x)= \log_{4}(1-x)+\log_{0.25}(2x+1)\]

OpenStudy (anonymous):

x = 0 or 7

OpenStudy (anonymous):

solution please!!

OpenStudy (anonymous):

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

OpenStudy (shubhamsrg):

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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