How do you solve #10= 2e^ { 5x }#?

2 Answers
Jim G.
Feb 5, 2018

#x~~0.322" to 3 dec. places"#

Explanation:

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

#•color(white)(x)logx^nhArrnlogx#

#"divide both sides by 2"#

#rArre^(5x)=10/2=5#

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

#rArrlne^(5x)=ln5#

#rArr5xcancel(lne)^1=ln5#

#rArrx=ln5/5~~0.322" to 3 dec. places"#

Vansh T.
Feb 5, 2018

x=ln(5)/5

Explanation:

#10=2e^(5x)#
or #5=e^(5x)#
Take natural log on both sides
#ln(5)=ln(e^(5x))#

Using propertyenter image source here

#ln(5)=5x*ln(e)#
0r
ln(5)=5x [since ln(e)=1 ]

x=ln(5)/5

Related questions
Impact of this question
916 views around the world
You can reuse this answer
Creative Commons License