Question

Solve In 2 + In x=5. Round to the nearest thousandth, if necessary. Can you show the steps so I can go off this one for my other problems like this

Answer 1

ln rules
product rule \[\huge\rm \ln x + \ln y = \ln (x \times y)\]
if there is plus sign you can change that to multiplication

Answer 2

So I just multiply?

Answer 3

so how would you change this \[\huge\rm ln 2 + \ln x = ??\]
can be written as ??

Answer 4

In(2 x x) ??

Answer 5

yep right \[\huge\rm ln (2x) = 5\]
now you have to take "e" both side to cancel out ln
for example

Answer 6

\[\huge\rm e^{\ln +1} = 1\]
e can cancel out ln

Answer 7

Oh goodness.. Im getting confused, if you did that would the In just cancel out? or

Answer 8

yes right left side ln should cancel out

Answer 9

okay Im following now

Answer 10

yep right \[\huge\rm e^{ln (2x)} = e^5\]
now left side e^{ln} can cancel out

Answer 11

alright, then do I replace the In with e^5?

Answer 12

nope that stays same

Answer 13

oh okay

Answer 14

when you cancel out e^{ln} then you should have yep right \[\huge\rm 2x = e^5\]

now divide both side by 2

Answer 15

I always get stuck on this step, so x=

Answer 16

just divide both side 2 \[\huge\rm \frac{ 2x }{ 2 }=\frac{ e^5 }{ 2 }\]

Answer 17

Thats what I did but I wasnt sure if it was right

Answer 18

it says "round to the nearest thousandth"
so use calculator put e^5/2

Answer 19

74.21?

Answer 20

thousandth

Answer 21

\[\huge\rm \frac{ e^5 }{ 2 } =74.2065795513\]
now round to the nearest thousandth

Answer 22

+3 decimal places

Answer 23

74.206

Answer 24

5 is nexxt to the zero so
6 suppose to be