solving ln property equations such as: $\ln\sqrt{b^{−2}c^{−3}a^{1}}$

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Hi I have a homework problem that is always wrong and I dont know which problem I have wrong. Its really annoying since it doesn't indicate which one is wrong, it just says something is incorrect. I have gone through these multiple times and still can't figure it out.

If $\ln a=2$, $\ln b=3$, and $\ln c=5$, evaluate the following:

(a) $$\ln\frac{(a^{3})}{(b^{−3}c^{−4})} = -23$$

(b) $$\ln\sqrt{b^{−2}c^{−3}a^{1}}= -19/2$$

(c) $$\frac{\ln(a^{−1}b^{2})}{ln(bc)^{−1}}= 12$$

(d) $${\ln c^{−2}}{{(\ln\frac{a}{b^{2}}})^{3}}= 640$$

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There are 2 best solutions below

1
On

ln a=2, ln b=3, and ln c=5

$\ln\frac{a^3}{(b^{−3})(c^{−4})} = \ln a^3 - \ln ((b^{−3})(c^{−4})) = \ln a^3 -(\ln b^{−3} + \ln c^{−4})= \ln a^3 -\ln b^{−3} - \ln c^{−4} = 3\ln a + 3 \ln b + 4 \ln c = 6 + 9 +20 = 35$
so that one is wrong.

$\ln ((b^{−2})(c^{−3})(a^1))^{1/2} = \frac 12 [-2 \ln b - 3 \ln c + \ln a]= \frac 12 [-6 - 15 + 2] = -19/2$ so that one is right.

3
On

Look at the exponents on (a). You'll find at least part of your issue there. (c) also appears to be incorrect, but it's not super clear what the equation should be. Should the (-1) exponent in the denominator apply to only the (bc) or to the entire logarithm?