0;x \neq 1;x \neq 16;x \neq 64
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\cfrac{1}{log_{2}x} \cdot \cfrac{1}{log_{2} \frac{x}{16}} =\cfrac{1}{log_{2} \frac{x}{64}}
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\frac{1}{log_{2}x} \cdot \frac{1}{log_{2}-log_{2}16} =\frac{1}{log_{2}-log_{2}64}
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\frac{1}{log_{2}x} \cdot \frac{1}{log_{2}-4} =\frac{1}{log_{2}-6} " alt="(log_{x}2)(log_{ \frac{x}{16}}2)=log_{ \frac{x}{64} }2
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x>0;x \neq 1;x \neq 16;x \neq 64
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\cfrac{1}{log_{2}x} \cdot \cfrac{1}{log_{2} \frac{x}{16}} =\cfrac{1}{log_{2} \frac{x}{64}}
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\frac{1}{log_{2}x} \cdot \frac{1}{log_{2}-log_{2}16} =\frac{1}{log_{2}-log_{2}64}
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\frac{1}{log_{2}x} \cdot \frac{1}{log_{2}-4} =\frac{1}{log_{2}-6} " align="absmiddle" class="latex-formula">
![log_{2}x=a
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\frac{1}{a} \cdot \frac{1}{a-4} =\frac{1}{a-6}
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\frac{1}{a(a-4)} =\frac{1}{a-6}
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a(a-4)=a-6
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a^2-4a-a+6=0
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a^2-5a+6=0
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(a-2)(a-3)=0
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a_1=2
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log_{2}x_1=2
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x_1=2^2=4
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a_2=3
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log_{2}x_2=3
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x_2=2^3=8](https://tex.z-dn.net/?f=log_%7B2%7Dx%3Da%0A%5C%5C%5C%0A+%5Cfrac%7B1%7D%7Ba%7D+%5Ccdot+%5Cfrac%7B1%7D%7Ba-4%7D+%3D%5Cfrac%7B1%7D%7Ba-6%7D+%0A%5C%5C%5C%0A%5Cfrac%7B1%7D%7Ba%28a-4%29%7D+%3D%5Cfrac%7B1%7D%7Ba-6%7D+%0A%5C%5C%5C%0Aa%28a-4%29%3Da-6%0A%5C%5C%5C%0Aa%5E2-4a-a%2B6%3D0%0A%5C%5C%5C%0Aa%5E2-5a%2B6%3D0%0A%5C%5C%5C%0A%28a-2%29%28a-3%29%3D0%0A%5C%5C%5C%0Aa_1%3D2%0A%5C%5C%5C%0Alog_%7B2%7Dx_1%3D2%0A%5C%5C%5C%0Ax_1%3D2%5E2%3D4%0A%5C%5C%5C%0Aa_2%3D3%0A%5C%5C%5C%0Alog_%7B2%7Dx_2%3D3%0A%5C%5C%5C%0Ax_2%3D2%5E3%3D8 "log_{2}x=a
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\frac{1}{a} \cdot \frac{1}{a-4} =\frac{1}{a-6}
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\frac{1}{a(a-4)} =\frac{1}{a-6}
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a(a-4)=a-6
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a^2-4a-a+6=0
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a^2-5a+6=0
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(a-2)(a-3)=0
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a_1=2
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log_{2}x_1=2
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x_1=2^2=4
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a_2=3
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log_{2}x_2=3
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x_2=2^3=8")
Ответ: 4; 8