# Logarithm

• Mar 31st 2008, 12:26 PM
chrozer
Logarithm
How would you find the values of $x$ for these logarithms:

a. $log_x \frac {1}{5} = -\frac {1}{4}$
b. $log_x 81 = \frac {3}{4}$

I put both of them in exponential form:

a. $x^\frac {-1}{4} = \frac {1}{5}$
b. $x^\frac {4}{3} = 81$

Then I got stuck in trying to figure out what the value of "x" equals. I knw how to find it using a calculator but I can't find the value without using a calculator.

And one more thing. How do you find the inverse fuction of $f(x)=2log_2 x$?
• Mar 31st 2008, 12:45 PM
TheEmptySet
Quote:

Originally Posted by chrozer
How would you find the values of $x$ for these logarithms:

a. $log_x \frac {1}{5} = -\frac {1}{4}$
b. $log_x 81 = \frac {3}{4}$

I put both of them in exponential form:

a. $x^\frac {-1}{4} = \frac {1}{5}$
b. $x^\frac {4}{3} = 81$

Then I got stuck in trying to figure out what the value of "x" equals. I knw how to find it using a calculator but I can't find the value without using a calculator.

And one more thing. How do you find the inverse fuction of $f(x)=2log_2 x$?

for a) use exponent laws and raise each side to -4th power

$(x^\frac {-1}{4})^{-4} = (\frac {1}{5})^{-4} \iff x=5^4=625$

for b.

$x^{4/3}=81 \iff x^{4/3}=3^4$ raise each side to the 3/4th power

$(x^{4/3})^{3/4}=(3^4)^{3/4} \iff x=3^3=27$

Yeah!!(Rock)
• Mar 31st 2008, 12:49 PM
TheEmptySet
forgot the last one... :(
$
f(x)=2log_2 x
$

$y=2\log_2(x)$ exchange x and y and solve for y

$x=2\log_2(y) \iff \frac{x}{2}=\log_2(y) \iff 2^{\frac{x}{2}}=y$

So the inverse is

$f^{-1}(x)=2^{\frac{x}{2}}$
• Mar 31st 2008, 12:49 PM
chrozer
Quote:

Originally Posted by TheEmptySet
for a) use exponent laws and raise each side to -4th power

$(x^\frac {-1}{4})^{-4} = (\frac {1}{5})^{-4} \iff x=5^4=625$

for b.

$x^{4/3}=81 \iff x^{4/3}=3^4$ raise each side to the 3/4th power

$(x^{4/3})^{3/4}=(3^4)^{3/4} \iff x=3^3=27$

Yeah!!(Rock)

Thanks alot! One more thing - How do you find the inverse of $f(x)=2log_2 x$?

EDIT- Ok I see now. Thnx very much!