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# Log questions

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1. Let x and b be positive real numbers so that \log_b(x^2) = 10. Find \log_{\sqrt[3]{b}} \left( \frac{1}{x} \right). 2. Let a, b, and c be the roots of 24x^3 - 121x^2 + 87x - 8 = 0. Find \log_3(a)+\log_3(b)+\log_3(c). I'm not really sure how to approach these ones. The best I did was to try to use exponential notation but that didn't seem to work out so well. All help is appreciated!
Mar 13, 2023

#1
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1.

We can start by using the properties of logarithms to simplify log_b x^2:

log_b x^2 = 10 2 log_b x = 10 log_b x = 5

Now we can use this result to simplify log_b^(1/3) 1/x:

log_b^(1/3) 1/x = (log_b (1/x))^(1/3)

Next, we can use the property that log_b (1/x) = -log_b x:

log_b^(1/3) 1/x = (-log_b x)^(1/3)

Substituting in the value we found earlier for log_b x:

log_b^(1/3) 1/x = (-5)^(1/3)

We can simplify this using the fact that (-a)^(1/3) = - (a^(1/3)) for any real number a:

log_b^(1/3) 1/x = - (5^(1/3))

So the final answer is -5^(1/3)

Mar 13, 2023
#2
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2. By Vieta's formulas, a + b + c = 121/24.

Then log_3 (a) + log_3 (b) + log_3 (c) = log_3 (121/24).  You can simplify this a bit, by writing it as

log_3 (121/24) = log_3 (121//(3*8)) = log_3 (121/8) - log_3 3 = log_3 (121/8) - 1.

Guest Mar 13, 2023
#3
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Please just put one question per post.

It is amazing that you got any answer at all.  Your questions are all run together and barely readable.

This is how they shouold have been presented, only in 2 different posts.

1. Let x and b be positive real numbers so that     $$\log_b(x^2) = 10$$.

Find $$\log_{\sqrt[3]{b}} \left( \frac{1}{x} \right).$$

2. Let a, b, and c be the roots of        $$24x^3 - 121x^2 + 87x - 8 = 0$$.

Find           $$\log_3(a)+\log_3(b)+\log_3(c)$$.

Mar 13, 2023