SOLUTION: log(base)2 x+log(base)8 x+log(base)64 x=3 (then value x)
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Question 763915: log(base)2 x+log(base)8 x+log(base)64 x=3 (then value x)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You need to use the change of base formula:
\ =\ \frac{\log_b(x)}{\log_b(a)})
\ =\ \frac{\log_2(x)}{\log_2(8)})
But then, since  \ \ \Leftrightarrow\ \ b^y = x)
 \ \ \Leftrightarrow\ \ 2^y = 8)
hence 
, so
\ =\ \frac{\log_2(x)}{3})
Similarly
\ =\ \frac{\log_2(x)}{\log_2(64)}\ =\ \frac{\log_2(x)}{6})
Substituting
\ +\ \frac{\log_2(x)}{3}\ +\ \frac{\log_2(x)}{6}\ =\ 3)
Factor out )
\left(1\ +\ \frac{1}{3}\ +\ \frac{1}{6}\right)\ =\ 3)
A little arithmetic, and
\ =\ 2)
But since
\ =\ 2\ \Rightarrow\ x\ =\ 4)
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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