# Test: Logarithms II - Ambitious

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Question 1:   Calculate ${25}^{{\mathrm{log}}_{5}\left(4\right)}+{e}^{2-\mathrm{ln}\left({e}^{2}\right)}+{3}^{{\mathrm{log}}_{5}\left(5\right)}=...$
$20$
$16$
$1$
$8$
Question 2:   Find $x$, if ${x}^{{\mathrm{log}}_{4}\left(16\right)}=25$
${x}_{1}=5,\text{\hspace{0.17em}}{x}_{2}=-5$
$x=\frac{5}{2}$
${x}_{1}=2,\text{\hspace{0.17em}}{x}_{2}=-2$
$x=4$
Question 3:   Calculate ${\mathrm{log}}_{5}\left(48\right)$, if ${\mathrm{log}}_{5}\left(3\right)=a,\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(2\right)=b$
$a+b$
$a+4b$
$2a+4b$
${a}^{2}{b}^{2}$
Question 4:   Find $x$, if ${\mathrm{log}}_{\frac{1}{6}}\left({x}^{2}-7x+18\right)=-1$
${x}_{1}=-3,\text{\hspace{0.17em}}{x}_{2}=4$
${x}_{1}=3,\text{\hspace{0.17em}}{x}_{2}=4$
$x=0$
$x=1$
Question 5:   Find $x$, if ${\mathrm{log}}_{4}\left(\frac{x}{4096}\right)×{\mathrm{log}}_{4}\left(x\right)=7$
$x=32$
${x}_{1}=256,\text{\hspace{0.17em}}{x}_{2}=16$
${x}_{1}=16384,\text{\hspace{0.17em}}{x}_{2}=\frac{1}{4}$
${x}_{1}=4096,\text{\hspace{0.17em}}{x}_{2}=8$
Question 6:   Calculate ${\mathrm{log}}_{5}\left(0.04\right)+{\mathrm{log}}_{5}\left(25\right)=...$
$0$
$1$
$0.04$
$5$
Question 7:   Calculate ${5}^{3{\mathrm{log}}_{5}\left(2\right)}+{\mathrm{log}}_{\frac{1}{4}}\left(32\right)=...$
$5.5$
$2$
$0.5$
$-5.5$
Question 8:   Find $x$, if $2{\mathrm{log}}_{5}\left(x\right)+{\mathrm{log}}_{5}\left(4\right)=2$
$x=4$
$x=1$
$x=±\frac{5}{2}$
$x=\frac{5}{2}$
Question 9:   Find $x$, if ${\mathrm{log}}_{2}^{2}\left(x\right)=9$
$x=\frac{1}{8}$
${x}_{1}=3,\text{\hspace{0.17em}}{x}_{2}=\frac{1}{3}$
${x}_{1}=8,\text{\hspace{0.17em}}{x}_{2}=\frac{1}{8}$
$x=8$
Question 10:   Find $x$, if ${\mathrm{log}}_{5}\left({x}^{2}-3x+7\right)=5$
${x}_{1}=-2,\text{\hspace{0.17em}}{x}_{2}=1$
${x}_{1}=2,\text{\hspace{0.17em}}{x}_{2}=1$
$x=0$
$x=1$