solver
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- May 22, 2009
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Can any explain help me solve this question:
a^{2}+ b^2=7ab Prove that log{x}{a+b}over{3}={1}over{2}(log{x}a+log{x}b)
a^{2}+ b^2=7ab Prove that log{x}{a+b}over{3}={1}over{2}(log{x}a+log{x}b)
Logarithm relation
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