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Algebra
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ln 6 + ln x = 1.3
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solve for x by changing to exponential form. ln(6) + ln(x) = 1.3
recall your log rules \(\bf ln({\color{blue}{ 6}}) + ln({\color{blue}{ x}}) = 1.3\implies ln({\color{blue}{ 6\cdot x}})=1.3\implies \large log_{\color{red}{ e}}({\color{blue}{ 6\cdot x}})=1.3 \to ?\)
yes, I have that I=on my paper. I just don't know where to go from there. would I divide 6? so if I do that it would log x = .217
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