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How do you solve #4( e ^ { x } + 1) = 12#?

2 Answers
Mar 8, 2018

Answer:

#x=0.693#, to three d.p.

Explanation:

We have #4(e^x+1)=12#

So:

#e^x+1=3#

#e^x=2#

#lne^x=ln2#

#xlne=ln2#

#x=ln2#

#x=0.693#

Mar 8, 2018

Answer:

#x=ln2#

Explanation:

#"divide both sides by 4"#

#rArre^x+1=3#

#rArre^x=3-1=2#

#•color(white)(x)[logx^nhArrnlogx" and "lne^x=x]#

#"take ln of both sides"#

#lne^x=ln2#

#rArrx=ln2#