Question

Solve for x

Answer 1

4e^3x-1=5
Solve using natural logarithms

Answer 2

i know the answer is 0.41 but i dont understand how to get it.

Answer 3

\[\ln 4e ^{3x-1}=\ln 5\]

Answer 4

then
\[\ln 4+3x-1=\ln 5\]

Answer 5

\[3x=\ln5-\ln4+1\]

Answer 6

and then u divide everything on the other side by the 3

Answer 7

right?

Answer 8

yeah right :)

Answer 9

I got it! :D Thanks man!

Answer 10

only question is, is how you got from the original equation to \[\ln 4e ^{3x-1}=5\]

Answer 11

\( \bf 4e^{3x-1}=5\qquad \textit{log cancellation rule of }{\color{blue}{ log_aa^x=x}}\qquad thus
\\ \quad \\
4e^{3x-1}=5\implies log_{\color{red}{ e}}{\color{red}{ e}}^{3x-1}=log_{\color{red}{ e}}5\implies 3x-1=log_{\color{red}{ e}}5\)

Answer 12

\(\large log_{{\color{red}{ whatever}}}{\color{red}{ whatever}}^x=x\)

Answer 13

:| I'm so lost now.

Answer 14

hehe

Answer 15

if you use a log, whose base MATCHES the value inside it
the result will be the exponent of the value

Answer 16

if you have a VALUE of "e", then you'd use a log with base "e", or ln
if you had say something like \(\bf 45^{x+3}=5\implies log_{{\color{red}{ 45}}}({\color{red}{ 45}}^{x+3)}=5\implies x+3=5\)

Answer 17

so \[\log_{15} \] is \[\log_{^{15}} \] ?????

Answer 18

woops... darn typo \(\bf 45^{x+3}=5\implies log_{{\color{red}{ 45}}}({\color{red}{ 45}}^{x+3)}=log_{{\color{red}{ 45}}}5\implies x+3=log_{{\color{red}{ 45}}}5\) rather

Answer 19

\(log_{15}?\)

Answer 20

what do you mean?

Answer 21

nvm that lol, i think i get it. So if there is a whole number(45) before the equation then when you put in into log form then it(45) will be the base and the number after the base?

Answer 22

and vice versa?

Answer 23

yes

Answer 24

so if it was \[\log_{12} 12^{3x-1}\]
then the exponential form would be
12\[^{3x-1}\]

Answer 25

\[12^{3x-1}\]

Answer 26

is what i ment to write.

Answer 27

\(\large {\bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}
\\ \quad \\ \quad \\
log_{12} 12^{3x-1}=\square \implies 12^{\color{red}{ \square}} =12^{{\color{red}{ 3x-1}}}}\)

Answer 28

im really confused now :| But i figured out what i was doing wrong in the first place. i divided ln(5/4) by 3, and then added 1 and that messed it up.

Answer 29

\(\large{ \bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}
\\ \quad \\ \quad \\
log_{{\color{red}{ 12}}} 12^{3x-1}=\square \implies {\color{red}{ 12}}^{\color{blue}{ \square}} =12^{{\color{blue}{ 3x-1}}}
\\ \quad \\
\implies \square =3x-1\qquad thus\qquad log_{{\color{red}{ 12}}} 12^{3x-1}=3x-1}\)

Answer 30

recall that a log function returns an exponent, the exponent needed to raise the log base, to obtain the value in it

Answer 31

that formula helps a ton. I think i get it a little better now.
i didnt realize what it was till now. so basicly
if it was, \[\log_{4} 6=y\]
then using the formula i would get
\[4^{y}=6\]

Answer 32

yeap

Answer 33

Ok i get it now! Thanks again :D