Question

if ln(a)=2 ln(b)=3 ln(c)=5 evaluate the following:
1) ln((a^-4)/(b^-4c^-3))
2) (ln c^-4)(ln(a/b^3))^4

Answer 1

http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif <--- use rule 2 listed there, then rule 3 then rule 1 for \(\bf ln\left(\cfrac{a^{-4}}{b^{-4}c^{-3}}\right)\)

for \(\bf \left[\cfrac{ln(c^{-4})}{ln\left(\frac{a}{b^3}\right)}\right]^4\)

distribute the exponent first
then use rule 2 listed there, then rule 2 again

Answer 2

Just as @jdoe0001 said follow this rule and solve the problem. If u r stuck somewhere please do come back. we are ready to help ya @homeworkgeek

Answer 3

I am not too sure exactly where I'm stuck at, but I keep getting the wrong answer even after following the log rules. not really sure what im missing @heybhai

Answer 4

Please elaborate the point where u are stucked @homeworkgeek

Answer 5

okay so this is how I attempted the first problem:
\[\ln (\frac{ a ^{-4} }{ b ^{-4} c ^{-3}})\]
\[\ln (a ^{-4})-\ln (b ^{-4})+\ln(c ^{-3})\]
\[-4\ln (a)+4\ln (b)-3\ln (c)\]
\[-4(2)+4(3)-3(5)\]
\[-8+12-15\]
-11

Answer 6

It will be
ln(a^−4)−ln(b^−4)-ln(c^−3) Note that in third term it will be -ve sign as it has come from denominator
-4ln(a)+4ln(b)+3ln(c)
-4(2)+4(3)+3(5)
-8+12+15
19

Answer 7

Thank you! that makes a lot more sense!

Answer 8

Do mind of change in sign when transposing number from numerator to denominator and vice-versa and do follow the link as @jdoe0001 suggested

Answer 9

\(\bf ln\left(\cfrac{a^{-4}}{b^{-4}c^{-3}}\right)
\\ \quad \\ \quad \\
ln(a^{-4})-[ln(b^{-4}c^{-3})]\implies ln(a^{-4})-[{\color{blue}{ ln(b^{-4})+ln(c^{-3})}}]
\\ \quad \\
ln(a^{-4}){\color{blue}{ -ln(b^{-4})-ln(c^{-3})}}\implies -4ln(a)-[-4ln(b)]-[-3ln(c)]
\\ \quad \\
-4ln(a)+4ln(b)+3ln(c)\)