a)

${\log }_3\left(x-5\right)=3$  

Założenia:

$x-5>0$  

$x>5$  

Z definicji logarytmu:

$3^3=x-5$  

$27=x-5\mid +5$  

$32=x$  

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b)

${\log }_{\frac{2}{3}}\left(x+2\right)=-2$  

Założenia:

$x+2>0$  

$x>-2$  

Z definicji logarytmu:

${\left(\frac{2}{3}\right)}^{-2}=x+2$  

${\left(\frac{3}{2}\right)}^2=x+2$  

$\frac{9}{4}=x+2\mid -2$  

$\frac{1}{4}=x$  

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c)

${\log }_{0,1}\left(2x-3\right)=1$  

Założenia:

$2x-3>0$  

$2x>3$  

$x>1,5$  

Z definicji logarytmu:

${\left(0,1\right)}^1=2x-3$  

$0,1=2x-3\mid +3$  

$3,1=2x\mid :2$  

$1,55=x$  

 

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d)

${\log }_2\left(3+\frac{x}{2}\right)=-4$  

Założenia:

$3+\frac{x}{2}>0$  

$6+x>0$  

$x>-6$  

Z definicji logarytmu:

$2^{-4}=3+\frac{x}{2}$  

${\left(\frac{1}{2}\right)}^4=3+\frac{x}{2}$  

$\frac{1}{16}=3+\frac{x}{2}\mid \cdot 16$  

$1=48+8x\mid -48$  

$-47=8x\mid :8$  

$-\frac{47}{8}=x$  

$-5\frac{7}{8}=x$  

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e)

${\log }_8\left(x^2-2\right)=-\frac{1}{3}$  

Założenia:

$x^2-2>0$  

$x^2>2$  

$x>\sqrt{2}\vee x<-\sqrt{2}$  

$x\in \left(-\infty ,\sqrt{2}\right)\cup \left(\sqrt{2}+\infty \right)$  

Z definicji logarytmu:

$8^{-\frac{1}{3}}=x^2-2$  

${\left(2^3\right)}^{-\frac{1}{3}}=x^2-2$  

$2^{3\cdot \left(-\frac{1}{3}\right)}=x^2-2$  

$2^{-1}=x^2-2$  

$\frac{1}{2}=x^2-2\mid +2$  

$\frac{5}{2}=x^2$  

Czyli:

$x=\sqrt{\frac{5}{2}}\vee x=-\sqrt{\frac{5}{2}}$  

$x=\frac{\sqrt{5}}{\sqrt{2}}\vee x=-\frac{\sqrt{5}}{\sqrt{2}}$  

$x=\frac{\sqrt{10}}{2}\vee x=-\frac{\sqrt{10}}{2}$  

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f)

${\log }_{\frac{1}{9}}\left(1-x^2\right)=\frac{3}{2}$  

Założenia:

$1-x^2>0$  

$\left(1-x\right)\left(1+x\right)>0$  

$x\in \left(-1,1\right)$  

Z definicji logarytmu:

${\left(\frac{1}{9}\right)}^{\frac{3}{2}}=1-x^2$  

${\left(\frac{1}{3^2}\right)}^{\frac{3}{2}}=1-x^2$  

${\left({\left(\frac{1}{3}\right)}^2\right)}^{\frac{3}{2}}=1-x^2$  

${\left(\frac{1}{3}\right)}^{2\cdot \frac{3}{2}}=1-x^2$  

${\left(\frac{1}{3}\right)}^3=1-x^2$  

$\frac{1}{27}=1-x^2\mid -1$  

$-\frac{26}{27}=-x^2\mid \cdot \left(-1\right)$  

$\frac{26}{27}=x^2$  

Czyli:

$x=\sqrt{\frac{26}{27}}\vee x=-\sqrt{\frac{26}{27}}$  

$x=\frac{\sqrt{26}}{\sqrt{27}}\vee x=-\frac{\sqrt{26}}{\sqrt{27}}$  

$x=\frac{\sqrt{26}}{3\sqrt{3}}\vee x=-\frac{\sqrt{26}}{3\sqrt{3}}$  

$x=\frac{\sqrt{78}}{9}\vee x=-\frac{\sqrt{78}}{9}$