How do you solve #(e^(x+5) / e^(5)) = 3#?

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Answer 1 · 2018-05-23T12:44:05

Solution: # x= 1.0986#

#e^(x+5)/e^5=3 or( e^x *cancel (e^5))/cancel(e^5)=3# or

#e^x =3# Taking natural log on both sides we get,

# x ln e = ln 3 or x = ln 3 [ln e=1] :. x ~~ 1.0986 (4 dp)#

Solution: # x= 1.0986# [Ans]

Answer 2 · 2018-06-08T03:24:07

#x~~1.10#

On the left side, we have the same bases, so we can subtract the exponents.

#(e^color(red)((x+5))/(e^color(blue)5))=e^(color(red)(x+5)-color(blue)5)=color(darkblue)(e^x)#

We now have the equation

#e^x=3#

The natural log (#ln#) function cancels with base-#e#, so we can take the natural log of both sides. We get

#cancel(ln)cancele^x=ln3#

#=>x=ln3#

#x~~1.10#

Hope this helps!