Suppose that f(x) = log (base 4) x
1. If f(3) = 0.79 and f(5) = 1.16, then evaluate f(75)
2. If f(3) is 0.79 find log (base 8) 3
3. Show that f(x) + f(1/x) = 0
THANKS GUYS!
You're expecetd to apply the various log rules you've been taught.
1. Note that $\displaystyle 75 = 3 \times 5^2$.
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2. $\displaystyle \log_8 3 = x \Rightarrow 8^x = 3 \Rightarrow (2^3)^x = 3 \Rightarrow 2^{3x} = 3 \Rightarrow \log_2 3 = 3x$ .... (1)
$\displaystyle \log_4 3 = 0.79 \Rightarrow 4^{0.79} = 3 \Rightarrow (2^2)^{0.79} = 3 \Rightarrow 2^{1.58} = 3 \Rightarrow \log_2 3 = 1.58$ .... (2)
Substitute (2) into (1) and solve for x.
There are undoubtedly easier ways of doing this but they're my first thoughts.
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3. Note that $\displaystyle f\left(\frac{1}{x}\right) = f(x^{-1}) = \log_4 x^{-1}$.
2.another method
$\displaystyle log_{8} 3=\frac{log_{4} 3}{log_{4}8}$
$\displaystyle =\frac{f(3)}{log_{4}{4} \times 2}$
$\displaystyle =\frac{f(3)}{(log_{4}{4}) + (log_{4}{ 2})}$
$\displaystyle =\frac{f(3)}{1+\frac{1}{log_{2}{4}}}$
$\displaystyle =\frac{f(3)}{1+\frac{1}{2}}$
continue...
originaly posted by goliath
given log base 4 3=.79 find log base 8 3. All parts answered Iadd a simplication for the latter calculation
log ( base 8) 8=1 log (base4 ) 8=3/2
therefore 3/2 is the factor to go from base 8 to base 4. going from base 4 to base 8 you multiply by 2/3
2/3log(base4)3=log (base8)3=2/3(,79)=.527
bjh