SOLUTION: a) write log(x^3 √(x+1)/(x-2)^2, x>2 as a sum and difference logarithms.Express all powers as factors. b) Find the exact value of the composite function cos[sin^-1(-1/3)]

Algebra.Com
Question 1189490: a) write log(x^3 √(x+1)/(x-2)^2, x>2 as a sum and difference logarithms.Express all powers as factors.
b) Find the exact value of the composite function cos[sin^-1(-1/3)]


Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
a) write , as a sum and difference logarithms.
Express all powers as factors.



=

=

=

=

True for all .
Verify solution : The solution is

b) Find the exact value of the composite function

let =>
This means that we are now looking for .
Next, use the identity :






Answer by MathTherapy(10553)   (Show Source): You can put this solution on YOUR website!

a) write log(x^3 √(x+1)/(x-2)^2, x>2 as a sum and difference logarithms.Express all powers as factors.
b) Find the exact value of the composite function cos[sin^-1(-1/3)]
a) 
    ------- Applying log-law:

RELATED QUESTIONS

(a) Write log a(a is base) √(x^2+1) /x^3(x+1)^4 , x > 0 as a sum and difference of... (answered by Alan3354)
write each expression as a sum and/or difference of logarithms. Express powers as... (answered by stanbon)
write each expression as a sum or difference of logarithms. express powers as factors. (answered by Abbey)
express as a sum or difference of logarithms. express powers as factors... (answered by Abbey)
write the following logarithms as a sum and/or difference of logarithms. Express powers... (answered by stanbon)
Write the expression as a sum and/or difference of logarithms. Express powers as factors. (answered by lwsshak3)
Write the expression as a sum and/or difference of logarithms. Express powers as factors. (answered by ankor@dixie-net.com)
ln((2x+3)/(x^2-3x+2))^2 Write expression as the difference of logarithms. Express... (answered by mathie123,Theo)
sample final questions: write as a single logarithm expression: have no clue how to... (answered by stanbon)