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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

$$\bullet \log {}_{\mathrm{c}}\left({\mathrm{a}}^{\mathrm{b}}\right)=\mathrm{b}{\log }_{\mathrm{c}}\left(\mathrm{a}\right)$$$$\bullet {\log }_{{\mathrm{c}}^{\mathrm{b}}}\left(\mathrm{a}\right)=\frac{1}{\mathrm{b}}{\log }_{\mathrm{c}}\left(\mathrm{a}\right)$$$$$$$$log{}_{0.5}\left(\sqrt{1+x}\right)+\frac{3}{2}{log}_{0.5}(1-x)=\frac{1}{4}log{}_{0.5}{(1-x^2)}^2+2$$$$log{}_{0.5}\left(\sqrt{1+x}\right)+{log}_{0.5}{(1-x)}^{\frac{3}{2}}=log{}_{0.5}{(1-x^2)}^{\frac{1}{2}}+2log{}_{0.5}(0.5)$$$$log{}_{0.5}\left(\sqrt{1+x}\right){(1-x)}^{\frac{3}{2}}=log{}_{0.5}{(1-x^2)}^{\frac{1}{2}}+log{}_{0.5}\left(\frac{1}{4}\right)$$$$log{}_{0.5}\left(\sqrt{1-x^2}\right)(1-x)=log{}_{0.5}\left(\frac{1}{4}\right){(1-x^2)}^{\frac{1}{2}}$$$$\left(\sqrt{1-x^2}\right)(1-x)=\left(\frac{1}{4}\right)\left(\sqrt{1-x^2}\right)$$$$1-x=\frac{1}{4}$$$$x=\frac{3}{4}$$$$$$$$If\left(\sqrt{1-x^2}\right)=0;$$$$1-x^2=0\to x=\pm 1$$$$butx=\pm 1isnotindomain.$$

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