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douglas12
Group Title
Use natural logarithms to solve the equation. Round to the nearest thousandth.
5e^2x+8=22
A. 0.515
B. 0.264
C. 0.896
D. –3.259
 one year ago
 one year ago
douglas12 Group Title
Use natural logarithms to solve the equation. Round to the nearest thousandth. 5e^2x+8=22 A. 0.515 B. 0.264 C. 0.896 D. –3.259
 one year ago
 one year ago

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dan815 Group TitleBest ResponseYou've already chosen the best response.0
what are you having trouble with
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
is the question \[5e^{2x + 8} = 22\] or \[5 e^{2x} + 8 = 22\]
 one year ago

douglas12 Group TitleBest ResponseYou've already chosen the best response.0
the second one
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
ok... so the 1st step is subtract 8 from both sides of the equation. then divide both sides of the equation by 5 what do you get?
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
dw:1373574672837:dw
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
you got it from there?
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
you must use ln to cancel e
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
ln means log of base e
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.0
so if you have log of base e of somethinh e to power something then the power must be your simplification
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
$$ \large { \text{log cancellation rule}\\ \color{blue}{log_aa^x = x}\\ e^{2x} = whatever\\ ln = log_e\\ log_e(e^{2x}) = log_e(whatever)\\ 2x = log_e(whatever) } $$
 one year ago
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