[tex] \bf \textit{Logarithm Cancellation Rules}
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\stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x
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\rule{34em}{0.25pt}\\\\
12\cdot 6^{t-1}=6\cdot e^{3t-1}
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12\cdot 6^t\cdot 6^{-1}=6\cdot e^{3t}\cdot e^{-1}\implies 12\cdot \cfrac{6^t}{6}=6\cdot \cfrac{e^{3t}}{e}\implies 2\cdot 6^t=6\cdot \cfrac{e^{3t}}{e} [/tex]
[tex] \bf \cfrac{2\cdot 6^t}{6}=\cfrac{e^{3t}}{e}\implies \cfrac{6^t}{3}=\cfrac{e^{3t}}{e}\implies e6^t=3e^{3t}\implies ln(e6^t)=ln(3e^{3t})
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ln(e)+ln(6^t)=ln(3)+ln(e^{3t})\implies 1+ln(6^t)=ln(3)+3t\cdot ln(e)
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1+t\cdot ln(6)=ln(3)+3t\implies t\cdot ln(6)-3t=ln(3)-1
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t[ln(6)-3]=ln(3)-1\implies t=\cfrac{ln(3)-1}{ln(6)-3}\implies t\approx -0.08161643824769 [/tex]
