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Equation: $\log \cos x(4) \times \log \cos^2 x(2) =1 $

*$\cos x, \cos^2 x$ - base
$4,2$ -numbers

$\cos x>0$ and not $1$ then $\cos x$ is in $(0;1)$

i've tried: if $\cos x=t$ then $\log t(4) \times \log t^2 (2) =1 ;$
$2\log t(2) \times \frac{1}{2} \log t(2)= \log t(t) /: \log t(2) ; $
$\log t(2) = \log t(t-2); $
$t=4 $

$\cos x=4$, i wonder if this is true since $\cos$ range is $[-1;1]$

Cheese Cake
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Ryuzaki
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    Yes, that's an equation...so? I bet you've already proved something, so please show your own work and effort. And what are those (4), (2) thingies? – DonAntonio Feb 09 '14 at 11:58
  • And $;1\neq\cos x>0\implies x\in\left(-\frac\pi2,,,\frac\pi2\right)\setminus{0};$ (principal value only) – DonAntonio Feb 09 '14 at 12:16

1 Answers1

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Justify the following by means of logarithmic properties:

$$1=\log_{\cos x}4\cdot\log_{\cos^2x}2=\log_{\cos x}4\frac{\log_{\cos x}2}{\log_{\cos x}\cos^2x}=\log_{\cos x}^22\implies$$

$$\log_{\cos x}2=\pm1\iff\cos^{\pm1}x=2\;,\;\;\text{so...}$$

You may be interested in the trigonometric equation

$$\sec x=\frac1{\cos x}=2\;\ldots$$

DonAntonio
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