Question

PLEASE HELP
Solve 1n 4 + 1n (3x) = 2 Round Your Answer To The Nearest Hundredth

Answer 1

Choices Are

A. 1.13
B. 0.18
C. 1.41
D. 0.62

Answer 2

here are some log rules to help you out :)
\[\ln a + \ln b = \ln ab\]
\[\ln a = \log_{e} a\]

Answer 3

So first, using the log laws above.
What would \[\ln 4 + \ln (3x) = ?\]

Answer 4

hint: 4 times 3x = ?

Answer 5

Um, I'm not sure

Answer 6

12x ?

Answer 7

yeah you got \[\ln(12x)=2\] to solve

Answer 8

I'm still confused?!

Answer 9

correct! :)
So,
\[\ln 4 + \ln 3x = 2 \]
\[\ln 12x = 2\]
use this rule to help you:
\[\ln a =\log_{e} a\]
and
\[\log_{e} a = b \]
\[e^{b}= a\]

Answer 10

I'm still not sure how to do it ??? :-/

Answer 11

12=e^2 ?

Answer 12

so first let's RE-WRITE "ln12x" in log form.
\[\ln12x=\log_{e}12x \]


So now we know:
ln12x=2

\[\log_{e} 12x = 2\]

Answer 13

(sorry i made a typo before LOL)
So, we want to use the rule to help solve for x:
\[\log_{e} a = b\] where \[a = e^{b}\]

apply this rule to \[\log_{e} 12x = 2\]

Answer 14

so, 12x = ?

Answer 15

2 ?

Answer 16

1n 4 + 1n (3x) = 2

by log rules:
ln ( 4*3x) = 2

now raise both sides to the power of e