log help

Oct 2009
8
0
\(\displaystyle \log_2 x + \log_4 x + \log_8 x = 11 \)

Completely lost as to how to start this... only things Ive noticed so far are that bases of each log are all exponents of 2.

Would appreciate the help understanding this problem :p

Err srry... When x =2; the log base 4 of x is half that of log base 2 of x; and the log base 8 of x is a third of the log base 2 of x.. Had it backwards!
 
Apr 2009
288
56
Change of base law (Changing from base \(\displaystyle a\) to base \(\displaystyle c\))
\(\displaystyle log_{a}b=\frac{log_{c}b}{log_{c}a}\)

and addition law
\(\displaystyle log_{a}b + log_{a}z=log_{a}bz\)

Helpful start
Using change of base law

\(\displaystyle log_{4}x=\frac{log_{2}x}{log_{2}4}\)
 
  • Like
Reactions: nmbala
Oct 2009
8
0
This leaves me kinda where i have been stuck at just from a slightly different approach...

\(\displaystyle \log_2 x + \frac{1}{2}\log_2 x + \frac{1}{3}\log_2 x =11 \)

Im sure I am forgetting some basic algebra rules; I treat \(\displaystyle \log_2 x \) as a variable and perform basic algebra getting \(\displaystyle \frac{11}{6}\log_2 x = 11 \) and then \(\displaystyle x = 64 \)

haha wow! I mustve made a calculator error earlier because redoing this in my head as i type it works out!

\(\displaystyle \log_2 64 = 6 \) Fowling my observations from earlier 6 + half of 6 + a 3rd of 6 = 11!

Thx for the help! The exponential relationship between bases is pretty cool! \(\displaystyle \log_2 x \) is always going to be twice that of \(\displaystyle \log_4 x \)