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1= -2b(lg((x+yb)/(ab)) - express the term ,,b" from this equation - ,,lg" is logarithm in base 10 - there is logarithm of a fraction
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i don't think you can isolate b algabraically
Really. The best I could come up with was \[b/(10^{1/2b})(x+by)=1/a\] or\[10^{1/2b}(x+by)/b=a\]
If you've to find b. Just substitute values of a,x and y and see
lg((b+xy)/ax)= 1/(-2x) --- from this equation how can be getting these results ? b=-xy+ax b=x(a-y) x=b/(a-y)
b=-xy+ax can't be possible. because if you substitute this in that equation: \[\log_{}((b+xy)/ax) = -1/(2x)\] \[\log_{}((-xy+ax+xy)/ax) = -1/(2x)\] \[\log_{}(ax/ax) = -1/(2x)\] or,\[0=-1/(2x)\] or\[x=\infty\]
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