How do you solve #2 = e^(5x)#?

3 Answers
sjc
Jun 5, 2018

#x=0.1386(4dp)#

Explanation:

#e^(5x)=2#

taking natural logs of both sides

#5x=ln2#

#=>x=1/5ln2#

#x=0.1386294361#

Jim G.
Jun 5, 2018

#x~~0.1386" to 4 dec. places"#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx^nhArrnlogx#

#•color(white)(x)log_b b=1#

#"take the ln (natural log ) of both sides"#

#ln2=lne^(5x)#

#5xcancel(lne)^1=ln2#

#x=ln2/5~~0.1386" to 4 dec. places"#

Jacobi J.
Jun 6, 2018

#x~~0.139#

Explanation:

#ln# (Natural Log) cancels with base-#e#, so we can take the natural log of both sides. We have

#ln2=cancel(lne)^(5x)#

#ln2=5x#

#x=ln2/5#

#x~~0.139#

Hope this helps!

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