• Aug 4th 2008, 02:32 AM
magentarita
Suppose that ln 2 = a and in 3 = b. Use rules of logarithms to write each log in terms of a and b.

(1) ln (2/3)

(2) ln (4throot{(2/3)}
• Aug 4th 2008, 02:35 AM
Chop Suey
$\displaystyle \ln{(a \cdot b)} = \ln{a} + \ln{b}$

$\displaystyle \ln{\left(\frac{a}{b}\right)} = \ln{a} - \ln{b}$

$\displaystyle \ln{(a)^n} = n\ln{a}$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$
• Aug 4th 2008, 03:07 AM
mr fantastic
Quote:

Originally Posted by Chop Suey
$\displaystyle \ln{(a \cdot b)} = \ln{a} + \ln{b}$

$\displaystyle \ln{\left(\frac{a}{b}\right)} = \ln{a} - \ln{b}$

$\displaystyle \ln{(a)^n} = n\ln{a}$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$

Since the pronumerals a and b are used in the question, your excellent advice might be misunderstood. Re-pronumeraling:

$\displaystyle \ln{(A \cdot B)} = \ln{A} + \ln{B}$

$\displaystyle \ln{\left(\frac{A}{B}\right)} = \ln{A} - \ln{B}$

$\displaystyle \ln{(A)^n} = n\ln{A}$

$\displaystyle \sqrt[n]{A} = A^{\frac{1}{n}}$
• Aug 4th 2008, 05:20 AM
magentarita
Good...
I thank both of you.