Question

Solving Exponential Equations Using logarithms

Answer 1

when I say log I mean log base 2
take log of both sides
log(2^(3x+1) = log(16)
3x+1=4
can you solve the rest?

Answer 2

I'm confused. how did you get 4? Do you know how to solve using natural log? @zzr0ck3r

Answer 3

you don't want to use natural log, log_2(16)
what this is saying is 2^? = 16

Answer 4

okay @zzr0ck3r

Answer 5

natural log for e^x

Answer 6

you can use it, but it makes it ugly
ln(2^(3x+1)) = ln(16)
(3x+1)*ln(2)=ln(16)
3x+1=ln(16)/ln(2)
x=ln(16)/(3*ln(2))-1/3
but
ln(16)/(3*ln(2))-1/3 = 1 and 1 is much prettier.

Answer 7

I understand everything except how 3x+1 becomes -1/3...@zzr0ck3r

Answer 8

x=(ln(16)/(3*ln(2))-1)/3 divide both sides by 3 after subtracting 1 from both sides

Answer 9

okay thank you for explaining it to me and for being patient @zzr0ck3r and @OneSecond

Answer 10

you really should use log base 2, that was the point of this question
is has 2^something on the left hand side, and 2^4 on the right hand side.
this SCREAMS log base 2:)

Answer 11

okay, I will try it.. thanks @zzr0ck3r

Answer 12

np