$ \log_a{b} \times \log_b{a} = $ ?
$ \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?
On
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 $$n \log_x{a} =\log_{x}{a^n}$$
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.