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log(base6)x + log(base6)10 = 1
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Have you considered using logarithm properties? log(a) + log(b) = log(ab) Stuff like that.
no, but i will try that.
And this one: \(log_{b}(a) = c\iff b^{c} = a\)
\[\log_6x+\log_610=1\]\[\log_6(10x)=1\]\[\log_6(10x)=\log_66\]\[10x=6~~~~~~x=3/5\]
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