Question

How do I solve this equation:

4500 = 1500(e)^(.04*x)

Answer 1

same like last one :-)
hint: \[\huge\rm ln e^x = x \]
natural log can cancel out e
bec ln is inverse of e

Answer 2

@Nnesha I divided by 1500 on both sides and got: 3 = e^(.04*x)
So now I do: lne^(.04*x)?

Answer 3

yep that's right :-)

Answer 4

@Nnesha I'm still confused on how I get the x out of the equation

Answer 5

so ln can cancel e
\[\huge\rm \cancel{ln e}^x = x ~~~ \cancel {ln e}^p = p\]

Answer 6

\[4500=1500e ^{0.4x}\]
This is a typical equation that combines logarithms and exponents, it is indeed a good way to put in practice those little algebraic properties.
Now, since the variable is in the exponent, we want to get rid of it, but first let's isolate the e^0.4x :
\[e ^{0.4x}=\frac{ 4500 }{ 1500 }\]
Now, in order to get that 0.4x to the numerator, we will take a natural logarithm on both sides:
\[\ln(e ^{0.4x})=\ln (\frac{ 4500 }{ 1500 })\]
and by logarithmic properties:
\[0.4x(lne)=\ln (\frac{ 4500 }{ 1500 })\]

Answer 7

to get rid of e you should always apply "ln" log
to get ride of "ln" you should always apply e

Answer 8

@Nnesha Sorry I'm still very confused. If I now have: ln(3) = ln(e)^(.04x) then do I cross out the ln on both sides and have 3 = e^(.04x) ? I feel like I'm going in circles

Answer 9

okay
"cross out the ln both sides"
nope you don't have to cross ln then question would be same l 3 = e^(.04x)
our purpose to take ln is to cancel out e and to bring the exponent down