log-0-5-1-x-3log-0-25-1-x-log-1-16-1-x-2-2-2-

Question Number 183594 by greougoury555 last updated on 27/Dec/22

\log_{0.5} \sqrt{1 + x} + 3 \log_{0.25} (1 - x) = \log_{1 / 16} {(1 - x^{2})}^{2} + 2

Answered by hmr last updated on 27/Dec/22

∙ \log_{c} (a^{b}) = b \log_{c} (a)

∙ \log_{c^{b}} (a) = \frac{1}{b} \log_{c} (a)

l o g_{0.5} (\sqrt{1 + x}) + \frac{3}{2} {l o g}_{0.5} (1 - x) = \frac{1}{4} l o g_{0.5} {(1 - x^{2})}^{2} + 2

l o g_{0.5} (\sqrt{1 + x}) + {l o g}_{0.5} {(1 - x)}^{\frac{3}{2}} = l o g_{0.5} {(1 - x^{2})}^{\frac{1}{2}} + 2 l o g_{0.5} (0.5)

l o g_{0.5} (\sqrt{1 + x}) {(1 - x)}^{\frac{3}{2}} = l o g_{0.5} {(1 - x^{2})}^{\frac{1}{2}} + l o g_{0.5} (\frac{1}{4})

l o g_{0.5} (\sqrt{1 - x^{2}}) (1 - x) = l o g_{0.5} (\frac{1}{4}) {(1 - x^{2})}^{\frac{1}{2}}

(\sqrt{1 - x^{2}}) (1 - x) = (\frac{1}{4}) (\sqrt{1 - x^{2}})

1 - x = \frac{1}{4}

x = \frac{3}{4}

I f (\sqrt{1 - x^{2}}) = 0;

1 - x^{2} = 0 \to x = \pm 1

b u t x = \pm 1 i s n o t i n d o m a i n .

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