Logarithms - simple but confusing

I can do the simple ones, but once they include fractions and stuff I get lost!

Using laws of logarithms, simplify the following expressions to a single log term:

a) 2/3 logx + 1/3 logxy^3

b) log (x^2 + 5x) - log (x+5)

c) log (x^2 - 64) - log (x+8)

10 points and a hug to whoever can help meee! thanks :)

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a) ⅔logx + ⅓log(xy³)
= ⅔logx + ⅓[ log(x) + log(y³)]
= ⅔logx + ⅓[ log(x) + 3log(y)]
= ⅔logx + ⅓log(x) + log(y)
= log(x) + log(y)


b) log(x² + 5x)   -   log(x+5)
= log[x(x + 5)]   -   log(x+5)
= log(x) + log(x + 5)   -   log(x+5)
= log(x)


c) log(x² - 64) - log(x+8)
= log(x² - 8²) - log(x+8)            ← notice the difference of squares x²-8²
= log[(x-8)(x+8)]   -   log(x+8)
= log(x-8) + log(x+8)   -   log(x+8)
= log(x-8)


Have a good one!!!

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Pretty easy ones.
a): 2/3 logx + 1/3 logxy^3=2/3 logx+1/3(logx+logy^3)=2/3 logx+1/3logx+1/3logy^3=logx+logy=log(xy)
b):log (x^2 + 5x) - log (x+5)=log(x(x+5))-log(x+5)=
logx+log(x+5)-log(x+5)=logx
c):log(x^2-64)-log(x+8)=log[(x+8)(x-8)]
-log(x+8)=
log(x+8)+log(x-8)-log(x+8)=log(x-8)

(a.) log(x^2/3) + log(x^1/3 * y) = log(xy)

(b.) log(x^2 + 5x/(x + 5)) = log(x)

(c.) log((x + 8)(x - 8)/(x + 8)) = log(x - 8)