Asked by Reina
log(base5)x+log(base25)x+log(base125)x=33
solve for x
solve for x
Answers
Answered by
Reiny
let's establish a property
let log(base25) x = y
25^y = x
5^(2y) = x
back to log notation ...
log(base5)x = 2y
y = (1/2)log(base5) x
so log(base5)x+log(base25)x+log(base125)x=33
log(base5)x + (1/2)log(base5)x + (1/3)log(base5)x = 33
(11/6)log(base5)x = 33
log(base5)x = 18
x = 5^18
let log(base25) x = y
25^y = x
5^(2y) = x
back to log notation ...
log(base5)x = 2y
y = (1/2)log(base5) x
so log(base5)x+log(base25)x+log(base125)x=33
log(base5)x + (1/2)log(base5)x + (1/3)log(base5)x = 33
(11/6)log(base5)x = 33
log(base5)x = 18
x = 5^18
Answered by
Anonymous
log base100 power1
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