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I'm rusty on logarithms. What is the approach to a problem like this? Any hints would be appreciated.

I'm thinking the subtraction on the numerator and denominator can become division since the bases are the same?

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  • $\begingroup$ Use $a \log_b m = \log_b m^a$ and $\log_b a - \log_b c =\log_b \frac ac$ to get that this is $\frac {\log_2\frac{24}{\sqrt {72}}}{\log_3 \frac {18}{\sqrt[3]{72}}}$. $\endgroup$
    – fleablood
    Jan 18, 2020 at 22:17

5 Answers 5

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The identities $\log_a b + \log_a c \equiv \log_a(bc)$ and $b\log_a c \equiv \log_a c^b$ will be needed here. You can then do \begin{align*} \frac{\log_2 24 - \frac 12 \log_2 72} {\log_3 18 - \frac 13 \log_3 72} &= \frac{\log_2 24 - \log_2 \sqrt{72}} {\log_3 18 - \log_3 \sqrt[3]{72}} \\ &= \frac{\log_2 \frac{24}{\sqrt{72}}} {\log_3 \frac{18}{\sqrt[3]{72}}} \\ &= \frac{\log_2 \frac{24}{3\sqrt 8}} {\log_3 \frac{18}{2\sqrt[3]{9}}} \\ &= \frac{\log_2 \frac{8}{\sqrt 8}} {\log_3 \frac{9}{\sqrt[3]{9}}} \\ &= \frac{\log_2 8 - \log_2 \sqrt 8} {\log_3 9 - \log_3 \sqrt[3]{9}} \\ &= \frac{\log_2 2^3 - \log_2 2^{3/2}} {\log_3 3^2 - \log_3 3^{2/3}} \\ &= \frac{3 - \frac 32} {2 - \frac 23} \quad \text{by definition} \\ &= \frac 98 \end{align*}

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    $\begingroup$ thank you for the detailed explanation! I am a little stuck on this: how does log(2)(8/sqrt(8)) become 3- 3/2? $\endgroup$
    – PineNuts0
    Jan 18, 2020 at 19:01
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    $\begingroup$ @PineNuts0 Sorry about that, I was a bit lazy and skipped a few steps. I have edited my answer somewhat, though let me know if it's still not clear enough about how you get the fractional exponents $\endgroup$ Jan 18, 2020 at 19:04
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    $\begingroup$ this is perfect ... thank you so much for taking time to spell out step by step. may I ask what software you use to type out the math? $\endgroup$
    – PineNuts0
    Jan 18, 2020 at 19:13
  • $\begingroup$ @PineNuts0 you're welcome :D. The way we type maths here is by directly entering LaTeX source. See this post for some more info. LaTeX is a very popular way to typeset maths, and well worth learning if you like your digital maths to look pretty. $\endgroup$ Jan 18, 2020 at 19:14
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1)

$\log_2 24 =\log_2 (3\times 2^3)= \log_2 3 + \log_2 2^3 = \log_2 3 + 3$.

$\frac 12\log_2 72 = \frac 12 (\log_2 3^2\times 2^3) = \frac 12(\log_2 3^2 + \log_2 2^3)=\frac 12(2\log 3 + 3) = \log 3 +\frac 32$

So $\log_2 24 -\frac 12\log_2 72 = (\log_2 3 + 3)-(\log_2 3 +\frac 32) = \frac 32$

Likewise.

$\log_3 18 = \log_3 2\times 3^2 = \log_3 2 + \log_3 3^2 = \log_3 2 + 2$

$\frac 13\log_3 72 = \frac 13(\log_3 3^2\times 2^3) = \frac 13(\log_3 3^2 + \log_3 2^3)=\frac 13(2 + 3\log_3 2)= \frac 23+ \log_3 2$.

So $\log_3 18-\frac 12\log_3 72 = (\log_3 2+2)-(\frac 23 + \log_3 2)= \frac 43$.

So $\frac {\log_2 24 -\frac 12\log_2 72}{\log_3 18-\frac 13\log_3 72}=\frac {\frac 32}{\frac 43}=\frac 32\frac 34=\frac {9}8$.

2)

$\frac {\log_2 24 -\frac 12\log_2 72}{\log_3 18-\frac 13\log_3 72}=$

$\frac {\log_2 24 -\log_2 72^{\frac 12}}{\log_3 18-\log_3 72^{\frac 13}}=$

$\frac {\log_2 \frac {24}{72^{\frac 12}}}{\log_3 \frac {18}{72^{\frac 13}}}=$

$\frac {\log_2 \frac {3*2^3}{3^{2*\frac 12}2^{3*\frac 12}}} {\log_3 \frac{2*3^2}{3^{2*\frac 13}2^{3*\frac 13}}}=$

$\frac {\log_2 2^{\frac 32}}{\log_3 3^{\frac 43}}=$

$\frac {\frac 32}{\frac 43}=\frac 32\frac 34=\frac 98$.

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Often better to factor the arguments of the logarithms to get to much simpler logarithms more quickly... \begin{align*} &\frac{\log_2 24 - \frac{1}{2} \log_2 72}{\log_3 18 - \frac{1}{3} \log_3 72} \\ &\quad{}= \frac{\log_2 (2^3 \cdot 3) - \frac{1}{2} \log_2 (2^3 \cdot 3^2)}{\log_3(2\cdot 3^2) - \frac{1}{3} \log_3 (2^3 \cdot 3^2)} \\ &\quad{}= \frac{\log_2 2^3 + \log_2 3 - \frac{1}{2} \left( \log_2 2^3 + \log_2 3^2 \right)}{\log_3 2 + \log_3 3^2 - \frac{1}{3} \left( \log_3 2^3 + \log_3 3^2 \right)} \\ &\quad{}= \frac{3\log_2 2 + \log_2 3 - \frac{1}{2} \left( 3\log_2 2 + 2\log_2 3 \right)}{\log_3 2 + 2\log_3 3 - \frac{1}{3} \left( 3\log_3 2 + 2\log_3 3 \right)} \\ &\quad{}= \frac{3 + \log_2 3 - \frac{1}{2} \left( 3 + 2\log_2 3 \right)}{\log_3 2 + 2 - \frac{1}{3} \left( 3\log_3 2 + 2 \right)} \\ &\quad{}= \frac{3 + \log_2 3 - \frac{3}{2} - \log_2 3 }{\log_3 2 + 2 - \log_3 2 - \frac{2}{3}} \\ &\quad{}= \frac{3 - \frac{3}{2}}{2 - \frac{2}{3}} \cdot \frac{6}{6} \\ &\quad{}= \frac{18-9}{12-4} \\ &\quad{}= \frac{9}{8} \text{.} \end{align*}

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Rule 1: $\ \log_a(x^q) = q\log_a(x)$

Rule 2 (Addition): $\ \log_a(x) + \log_a(y) = \log_a(xy), \ $

Rule 3 (Subtraction): $\ \log_a(x) - \log_a(y) = \log_a(\frac{x}{y}), \ $

Rule 3 is actually Rule 2 but replacing y with $\ \frac{1}{y} =y^{-1}$, and then using Rule 1.

Rule 4 (Change of base): $\ \frac{\log_a(x^p)}{\log_a(x^q)} = \log_q(p)$

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Yes, the subtractions can become division, and the fractions can become roots, plus $\log_a b =\frac 1 {\log_b a}$

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