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

Question

Caleb Alexander · Accepted Answer

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]

Madelyn Anderson · Answer

#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!

James Allen · Answer

undefined To solve the equation $\frac{{e^{(x+5)}}}{{e^5}} = 3$, you can begin by multiplying both sides of the equation by $e^5$ to eliminate the denominator:

$e^5 \times \frac{{e^{(x+5)}}}{{e^5}} = 3 \times e^5$

$e^{(x+5)} = 3e^5$

Next, take the natural logarithm (ln) of both sides to solve for $x$:

ln($e^{(x+5)}$) = ln($3e^5$)

$x + 5 = \ln(3e^5)$

Now, apply the properties of logarithms to simplify:

$x + 5 = \ln(3) + \ln(e^5)$

Remembering that $\ln(e^5) = 5$, we have:

$x + 5 = \ln(3) + 5$

Subtract 5 from both sides to isolate $x$:

$x = \ln(3)$