SOLUTION: I am suppose to solve this with out a calcutlator: {{{log(10,5sqrt(2))}}} the sqrt kinda throws me off, i have a tutor who tried to explain it but they were also confused on

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am suppose to solve this with out a calcutlator: {{{log(10,5sqrt(2))}}} the sqrt kinda throws me off, i have a tutor who tried to explain it but they were also confused on       Log On


   

Question 34336This question is from textbook intermediate algebra
: I am suppose to solve this with out a calcutlator:
log%2810%2C5sqrt%282%29%29
the sqrt kinda throws me off, i have a tutor who tried to explain it but they were also confused on what exactly to do. here is what we were trying:
log%2810%2C2%5E%281%2F5%29%29
log%2810%2C2%29%5E%281%2F5%29%2Blog%2810%2C1%29%5E%281%2F5%29
log10 cancles out
answere:
3%5E%281%2F5%29??????
we know we are going in the wrong direction with this but are very lost! please help! thanks! This question is from textbook intermediate algebra

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
I am suppose to solve this with out a calcutlator:
log%2810%2C5sqrt%282%29%29
FIRSTLY DO YOU MEAN 5*SQRT(2)....OR
FIFTH ROOT OF 2?
ASSUMING IT IS 5*SQRT(2)..DO AS FOLLOWS
1.FIND SQRT(2)=1.414..FROM CALCULATOR..STORE IT IN MEMORY
NOW FIND ITS LOG TO BASE 10 FROM CALCULATOR.RECALLING FROM MEMORY THE VALUE OF 1.414...YOU WILL GET THE ANSWER AS 0.1505
ON THE OTHER HAND IF IT IS FIFTH ROOT OF 2 ..
FIND 2^(1/5)=2^0.2.....USING X^Y FUNCTION IN THE CALCULATOR. YOU WILL GET
2^0.2=1.1487
AND LOG OF ABOVE NUMBER TO BASE 10 =0.0602
the sqrt kinda throws me off, i have a tutor who tried to explain it but they were also confused on what exactly to do. here is what we were trying:
log%2810%2C2%5E%281%2F5%29%29
log%2810%2C2%29%5E%281%2F5%29%2Blog%2810%2C1%29%5E%281%2F5%29
log10 cancles out
answere:
3%5E%281%2F5%29??????
we know we are going in the wrong direction with this but are very lost! please help! thanks!