1. ## Logs

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!

2. Originally Posted by goliath
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!
1. $f(75)=log_{4}75=log_{4}(5^2\times 3)=log_{4}(5^2)+log_{4} 3$
$=2log_{4}5+log_{4}3=2f(5)+f(3)$
now put values of f(5) and f(3)
2.
$log_{8} 3 =\frac{log_{2} {3}}{log_{2}{8}}=\frac{log_{2} {3}}{3log_{2}{2}} \ now \ continue$

3. Originally Posted by goliath
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 $75 = 3 \times 5^2$.

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2. $\log_8 3 = x \Rightarrow 8^x = 3 \Rightarrow (2^3)^x = 3 \Rightarrow 2^{3x} = 3 \Rightarrow \log_2 3 = 3x$ .... (1)

$\log_4 3 = 0.79 \Rightarrow 4^{0.79} = 3 \Rightarrow (2^2)^{0.79} = 3 \Rightarrow 2^{1.58} = 3 \Rightarrow \log_2 3 = 1.58$ .... (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 $f\left(\frac{1}{x}\right) = f(x^{-1}) = \log_4 x^{-1}$.

4. 2.another method
$log_{8} 3=\frac{log_{4} 3}{log_{4}8}$
$=\frac{f(3)}{log_{4}{4} \times 2}$
$=\frac{f(3)}{(log_{4}{4}) + (log_{4}{ 2})}$
$=\frac{f(3)}{1+\frac{1}{log_{2}{4}}}$
$=\frac{f(3)}{1+\frac{1}{2}}$
continue...

5. Originally Posted by mr fantastic
You're expecetd to apply the various log rules you've been taught.

1. Note that $75 = 3 \times 5^2$.

----------------------------------------------------------------------------

2. $\log_8 3 = x \Rightarrow 8^x = 3 \Rightarrow (2^3)^x = 3 \Rightarrow 2^{3x} = 3 \Rightarrow \log_2 3 = 3x$ .... (1)

$\log_4 3 = 0.79 \Rightarrow 4^{0.79} = 3 \Rightarrow (2^2)^{0.79} = 3 \Rightarrow 2^{1.58} = 3 \Rightarrow \log_2 3 = 1.58$ .... (2)

Substitute (2) into (1) and solve for x.

There are undoubtedly easier ways of doing this but they're my first thoughts.

-------------------------------------------------------------------------------

3. Note that $f\left(\frac{1}{x}\right) = f(x^{-1}) = \log_4 x^{-1}$.

Ok, for 3, where would I go after the inverse. I think it would be log_4 x + log_4 x^-1, but then what? Could you please explain? Thank you

6. Originally Posted by goliath
Ok, for 3, where would I go after the inverse. I think it would be log_4 x + log_4 x^-1, but then what? Could you please explain? Thank you
You're expected to know and be able to apply the log rule $\log_a b^m = m \log_a b$.

7. Originally Posted by mr fantastic
You're expected to know and be able to apply the log rule $\log_a b^m = m \log_a b$.

sweet, I got it. Now I have 1 more question

4. Find a function g such that f(g(x)) = 10f(x).

8. Originally Posted by goliath
sweet, I got it. Now I have 1 more question

4. Find a function g such that f(g(x)) = 10f(x).
$\log_4 g(x) = 10 \log_4 x$. Apply the rule I reminded you of and then draw a conclusion after comparing each side.

9. could I use g(x) = x^10 and then put that in as x and then put the 10 to the front of the log?

10. Originally Posted by goliath
could I use g(x) = x^10 and then put that in as x and then put the 10 to the front of the log?
$g(x) = 10^x$ is the correct answer. The End.

11. ## logs

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