How do you solve for x in

e^(8x+4)-9 = 0

?

Please show the steps, thanks :)

e^(8x+4)-9 = 0

e^(8x+4) = 9

Then you take the natural log of both sides( ln ) . You may not know what this means. the ln of a value gives out the value that you have to raise "e" to to get the value that you are operating on.
for example, ln(1) = 0, because e^ 0 = 1
ln(e) = 1 because e^1 = e
ln(e^6) = 6, because e^6 = e^6

note that ln (e^a) = a. (where a is anything. this is a property of logarithms) this is useful for our problem!

so, back to the problem in question

take the ln of both sides
ln( e^(8x+4) = ln(9) is the same as

8x+4 = ln(9)
8x = ln(9) -4
x = (ln(9)-4)/8

which is about -.225

hope this helps!

e^(8x+4)=9
Ln(e^(8x+4)=Ln9
8x+4=Ln9
x=((Ln9)-4)/8