Question

CALC II
I have a question. The question asks to solve for k for the equation: 3^(k/14)=a

We are using logarithms and exponentials but I just don't understand how to isolate k using that? please help!

Answer 1

logs undo exponents

Answer 2

log both sides

\[ln(3^{k/14})=ln(a)\]
exponents pull out of logs
\[(k/14)ln(3)=ln(a)\]
divide off the ln(3) and multiply thru my 14

Answer 3

i know so i use log or natural log (ln)? like do I use e^ln(k/14)??? like i dont get it for some reason

Answer 4

ohhhhh

Answer 5

I SEEE

Answer 6

i get it

Answer 7

thank you

Answer 8

cool

Answer 9

wow that was extremely stupid. I should have figured that out. thank you again! If i have any more questions I'll let you know!

Answer 10

good luck ;)

Answer 11

actually sorry I did type this into my webwork problem earlier and it said it was incorrect

Answer 12

Just a little interruption, but isn't this Pre-Calculus? :|

Answer 13

I put in 14[(ln(a))/(ln(3))] and it said incorrect

Answer 14

its a preview of exponential functions we are learning in calc II. so we know how to derive and integrate. anyway i dont get it

Answer 15

Another way to do it is,\[a^b = c \implies \log_a c = b\]\[3^{\large {k\over 14} }=a \implies \log_{3}a = {k \over 14} \implies 14\log_3 a = k\]

Answer 16

okayy i'll try that out! thank you

Answer 17

Well, the above still leads to the answer that @amistre64 provided, but still... :)

Answer 18

did I do the other way right though? it seems like it was correct. Im not sure why it says incorrect

Answer 19

you didnt simplify the results

Answer 20

\[k=14\frac{ln(a)}{ln(3)}\]
by change of base rule we get
\[k=14~log_3(a)\]

Answer 21

I also tried that answer to simplify it "14 log3(a)" and it still came out as incorrect
Im totally lost

Answer 22

are you sure you have the question correct?

Answer 23

yess it says solve for k in the equation 3exp (k/14)=a

Answer 24

\[3^{\frac{14log_3(a)}{14}}=a\]
\[3^{log_3(a)}=a\]
\[a=a\]

Answer 25

umm, exp(n) has a specific meaning; its means e^(n), eulers number

Answer 26

so I did it wrong? I just thought it meant to the power of

Answer 27

lets try it this way:\[3e^{k/14}=a\]
\[e^{k/14}=a/3\]
\[ln(e^{k/14})=ln(a/3)\]
\[(k/14)~ln(e)=ln(a/3)\]
\[k/14=ln(a/3)\]
\[k=14~ln(a/3)\]

Answer 28

ln(a/3) can also be written as: ln(a) - ln(3)

but i cant be sure which one is the "simplified" version of it

Answer 29

yes it's correct! thankyou so much. Now that I look at it, it's relatively simply lol. wow I feel so stupid but thank you!

Answer 30

being able to read the notation is useful ;) good luck

Answer 31

yeah If it had just said e^(whatever) I would have spent a lot less time on that lol