Question

example from book that i don't fully understand:

Answer 1

\[\log_{10} 2^{1.42}\]

Answer 2

according to this a theorem:
this is equal to; 1.42log2
=(1.42)(0.30103)
2^1.42=2.6758
how did they get the last part?

Answer 3

calculator ...

Answer 4

i can't use a calculator in the exam..

Answer 5

do you get log tables??

Answer 6

the book said use theorem 7 to compute \[2^{1.42}\]
and theorem 7 is: \[ \log_{a} X ^{y} = y \log_{a} x\]

Answer 7

nope only some logs like log2 and log3..

Answer 8

@lgbasallote

Answer 9

http://www.wolframalpha.com/input/?i=log+base+10+2^1.42
something fishy

Answer 10

i don't know but the answer for 2^1.42 = 2.6758

Answer 11

that's according to a calculator

Answer 12

we just need to solve it using the theorem.

Answer 13

it's not the answer to \[\log_{10} 2^{1.42}\]it's just telling you the value of \[2^{1.42}\]

Answer 14

maybe to compare

Answer 15

i know that \[\log_{10} 2^{1.42} = 0.42746\]
but we need to find 2^1.42

Answer 16

using logs

Answer 17

dint you just say \[2^{1.42} = 2.6758\]

Answer 18

anyway..you cant solve \[2^{1.42}\]without using a calculator...unless you know calculus or some other high level math

Answer 19

okay, i guess then i can't do my exercise.. thank you for all the help.

Answer 20

@waterineyes @satellite73

Answer 21

Which value you want to find :

\(2^{1.42}\)

Or:

\(log(2^{1.42})\) ??

Answer 22

the first using the second

Answer 23

And you are not allowed to use calculators ??

Answer 24

yes

Answer 25

Then I think you are allowed to Log Tables in exams...

Answer 26

maybe the book is wrong because i have said everything it has told me

Answer 27

Can you use Log Tables in exams or not ??

Answer 28

no log table but we are given certain logs that are needed

Answer 29

Can we use approximation here ??

Answer 30

i guess..