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  1. $ \log_a{b} \times \log_b{a} = $ ?

  2. $ \log_a{b} + \log_b{a} = \sqrt{29} $

What is $ \log_a{b} - \log_b{a} = $ ?

3.

What is b in the following:

$$ \log_b{3} + \log_b{11} + \log_b{61} = 1 $$

and

4.

$$ \frac{1}{log_2{x}} + \frac{1}{log_{25}{x}} - \frac{3}{\log_8{x}} = \frac{1}{\log_b{x}} $$ What is b?

Can anyone help me solve these?

colormegone
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ella
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2 Answers2

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The way to start all of these and turn them into simple algebra is that $\log_ab=\frac{\log_x b}{\log_x a}$ Using that formula, all of these become basic algebra. Give it a try and comment what you get.

Asimov
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  • $ \log_a{b} \times \log_b{a} = $ ?

    $\frac{\log_a}{\log_b}$ x $\frac{\log_b}{\log_a}$ and that equals 1 so that's what I got for the first question, but when I tried the second one I got stuck

     – ella Apr 04 '14 at 20:10
  • @JohnJPershing Please review the way you've written your expression: it is not really a good way of notating that... – colormegone Apr 04 '14 at 20:13
  • This is best written as $ \ \log_b x \ = \ \frac{\log_a x}{\log_a b} \ . $ So $ \ \log_b a \ = \ \frac{\log_a a}{\log_a b} \ . $ And that numerator is...? – colormegone Apr 04 '14 at 20:17
  • Thank you for your correction, but I'm still having trouble finding $ \log_a{b} - \log_b{a} = $ ? – ella Apr 04 '14 at 20:20
  • $\log_a b - \log_b a = \log_a b- \frac{log_a a}{log_a b}$ Does that help any more? – Asimov Apr 04 '14 at 20:27
  • I'm just not really sure how to use it with the square root of 29? – ella Apr 04 '14 at 21:44
  • make both sides a to the side, and the logs cancel, and the right becomes $a^\sqrt{29}$ – Asimov Apr 04 '14 at 21:46
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2 . let $ x=\log_a{b} =\frac{1}{ \log_b{a}} $ $$\log_a{b}+\frac{1}{\log_a{b}}=\sqrt{29}$$ Then $ \log_a{b}$ and $ \log_a{b}$ are solutions of the following equation

$$x+\frac{1}{x}=\sqrt{29}$$

$$x_1=\frac{\sqrt{29}-5}{2}=\log_b{a}$$ $$x_2=\frac{\sqrt{29}+5}{2}=\log_a{b}$$ (We assume that $a<b$ ) Thus $ \log_a{b}- \log_b{a}=\frac{\sqrt{29}+5}{2}-\frac{\sqrt{29}-5}{2}=5 $

4.

$$ \frac{1}{log_2{x}} + \frac{1}{log_{25}{x}} - \frac{3}{\log_8{x}} = \frac{1}{\log_b{x}} $$ Using formula $ \log_a{b} =\frac{1}{ \log_b{a}} $ we have $$ \log_x{2} + \log_{x}{25} - 3 log_x{8} = \log_x{b} $$ Use formulas $$\log_x{a} + \log_{x}{c}=\log_{x}{ac}$$ and $$n \log_x{a} =\log_{x}{a^n}$$

nadia-liza
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