Question

On a TI-84 Plus, how do you enter a log? like log4(bottom of log)^1

Answer 1

i knw about mine. ^.

Answer 2

^.^

Answer 3

which one do you have?

Answer 4

you mean you want the log base 4? like for example
\[\log_4(8)\]

Answer 5

Casio, fx-570ES

Answer 6

I think you need to use the change of base rule on the ti 84 plus

Answer 7

only does log base e or base 10

Answer 8

I like stroodle's approach, convert to base 10 or base e.

Answer 9

ok. I tried to put in log(1)4 and I got the right answer. However, when I did it with log(5)5, i did not.

Answer 10

it makes no difference really. fancy calculators have only the natural log these days. if you want
\[\log_4(8)\] put
\[\ln(8)\div \ln(6)\]

Answer 11

just looked at 84 + looks pretty straight forward. what exactly are you trying to compute?

Answer 12

ok, thank you so much!! Satellite73, we were trying to do just basic log math, but the answers from the professor did not match the ones we were getting.

Answer 13

ok well if you have a specific question we could walk through entering it into the calculator

Answer 14

log little 5 and then 5 with the answer being 1

Answer 15

well this is not a calculator problem for sure

Answer 16

\[\log_{5}5 \]

Answer 17

since you are basically being asked what power would you raise 5 to to get the number 5. the answer is clearly 1.

Answer 18

it takes longer to put this in the calculator than to think. BUT if you want to use a calculator hit
\[\ln(5)\div \ln(5)\] and you will get 1 for sure

Answer 19

idea is that if you want
\[\log_b(A)\] you can use
\[\log_b(A)=\ln(A)\div \ln(b)\]

Answer 20

ok. so i have to use ln instead of log on the cal?

Answer 21

it doesn't matter which one you use. so long as you don't mix them. you could use
\[log_b(A)=\log(A)\div\log(b)\] also

Answer 22

you have two logs on your calculator, log base ten which just say "log" and log base e which is written as "ln"

Answer 23

how did you know to divide the log5^5?

Answer 24

later on your mathematical experience you will not use log base ten which is just a relic of the fact that we write our numbers in base ten.

Answer 25

are you asking me why this works?

Answer 26

in other words, are you asking

\[\text{why is } \log_b(A)= \log(A)\div \log(b) ?\]

Answer 27

I guess my problem is when you have log4^1 and knowing which way to put it into the cal. but if u use ln (1)/ln (4) then it does not matter which way u put it in?

Answer 28

wait hold on. i am not sure what 'log4^1" means. do you mean
\[\log_4(1)\]?

Answer 29

again this is not a calculator problem. the statement
\[\log_4(1)=y\] means
\[4^y=1\]

Answer 30

yes, sorry about that!

Answer 31

ok so we just write this in equivalent exponential form and solve by thinking. we think "4 to what power is equal 1?" and the answer is 0 because
\[b^0=1\] for any b so
\[4^0=1\] so
\[\log_4(1)=0\]

Answer 32

in fact for any base b it is true that
\[\log_b(1)=0\] by the statement
\[b^0=1\]

Answer 33

i think you might be missing something. what does is
\[\log_3(9)\]? to find it i solve
\[3^y=9\] and since i know that
\[3^2=9\] i know
\[\log_3(9)=2\]

Answer 34

Wow. Ok, so that makes sense. It has been a few years since I had college algebra:) I used selective memory on this.

Answer 35

likewise to solve
\[\log_5(5)\] i solve
\[5^y=5\] and get y = 1

Answer 36

and to solve
\[\log_4(1)\] i solve
\[4^y=1\] and see that y = 0

Answer 37

so no calculator needed for these. now i can use it to cheat if i don't like thinking, or it if looks too complicated

Answer 38

for example if i want
\[\log_3(\frac{1}{81})\] i can solve
\[3^y=\frac{1}{81}\] and if i happen to know that \[3^{-4}=\frac{1}{81}\] i am done. but if i don't know that i can type in
\[\log(1\div 81)\div \log(3)\] and i will still get -4
you can try it

Answer 39

See, now that is my kind of thinking:)

Answer 40

yeah cheating is good if you don't feel like messing around with exponential notation. but you have to know how to do it. doesn't hurt to know some basic properties do you don't have to. in particular you should be fluent in going from
\[\log_b(A)=y\] to
\[b^y=A\] without hesitation

Answer 41

I guess I just have to go back and re-learn it. I am finished with math, but it is always nice to know. That makes more sense though! Thank u! Can I ask about this one? 7\[^{x}\]=12 the exponent is the x