Question

If log2=0.301 and log3=0.477, evaluate:

log 5??

Answer 1

log(a+b)=loga + logb
can you do it now?

Answer 2

not true. loga + logb = log(ab)

Answer 3

oh damn sorry.

Answer 4

euler is right

Answer 5

haha. i messed up. apologies.

Answer 6

log 5 it will use log(a/b) = loga - logv

Answer 7

log b

Answer 8

in my following solution i will imply that by log you mean log base 10 and not ln.

correct me if im wrong

Answer 9

yes you are right

Answer 10

haha ikr eyad. :/

Answer 11

wait, is it \(\large \log _{\color{red}{2}}x=0.301\)?????

Answer 12

you have to use that but

Answer 13

Then you can use what i call "flower power"

Answer 14

log 2 to the base 10 is 0.3..

Answer 15

ohhhh thank you @yrelhan4

Answer 16

ok what you need to do is..
log5= log(10/2) = log10 - log2..
log10=1 and you are given the balue of log 2.. solve.

Answer 17

\[y = \log_{b}x ---> x = b^{y}\]

so we are told that:

\[0.301 = \log_{10}2 \therefore 2 = 10^{0.301}\]
and
\[0.477 = \log_{10}3 \therefore 3 = 10^{0.477}\]

\[5 = 2+3 \therefore 5 = 10^{0.301} + 10^{0.477}\]

Answer 18

yrelhan is also correct, but i think my answer is the one he expected to do since he is given both informations.

both very valid methods though

Answer 19

no i need to use only logs of 2 and 3

Answer 20

Then follow what Euler said.

Answer 21

\[\therefore\] means therefore btw if you didn tknow

Answer 22

"flower power" => 2= 10^(0.301) haha

Answer 23

exactly what @Euler271 said