Question

Salut, ma puteti ajuta si pe mine la subpunctele b), d), e), f), g)?
Colegii imi spun ca ex sunt usoare, dar am lipsit la lectia ecuatii.

Answer 1

$b) {9}^{x} - 10 \times {3}^{x} + 9 = 0$

${( {3}^{2}) }^{x} - 10 \times {3}^{x} + 9 = 0$

${ ({3}^{x} )}^{2} - 10 \times {3}^{x} + 9 = 0$

${3}^{x} = t \: \: \: ,t > 0$

${t}^{2} - 10t + 9 = 0$

$a = 1$

$b = - 10$

$c = 9$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 10)}^{2} - 4 \times 1 \times 9$

$\Delta = 100 - 36$

$\Delta = 64$

$t_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 10) \pm \sqrt{64} }{2 \times 1} = \frac{10 \pm8}{2}$

$t_{1}= \frac{10 + 8}{2} = \frac{18}{2} = 9 = > {3}^{x} = 9 = > x_{1}=2$

$t_{2}= \frac{10 - 8}{2} = \frac{2}{2} = 1 = > {3}^{x} = 1 = > x_{2}=0$

$d)3 \times {16}^{x} + 2 \times {81}^{x} = 5 \times {36}^{x} \: | \div {16}^{x}$

$3 + 2 \times \frac{ {81}^{x} }{ {16}^{x} } = 5 \times \frac{ {36}^{x} }{ {16}^{x} }$

$3 + 2 \times {( \frac{81}{16}) }^{x} = 5 \times {( \frac{36}{16} )}^{x}$

$3 + 2 \times {( \frac{ {9}^{2} }{ {4}^{2} } )}^{x} = 5 \times {( \frac{9}{4} )}^{x}$

$3 + 2 \times{ [ {( \frac{9}{4} )}^{2} ] }^{x} = 5 \times {( \frac{9}{4}) }^{x}$

$3 + 2 \times {( \frac{9}{4} )}^{2 \times x} = 5 \times {( \frac{9}{4}) }^{x}$

$3 + 2 \times {( \frac{9}{4} )}^{2x} = 5 \times {( \frac{9}{4} )}^{x}$

$2 \times {( \frac{9}{4}) }^{2x} - 5 \times {( \frac{9}{4} )}^{x} + 3 = 0$

${( \frac{9}{4} )}^{x} = t \: \: \: ,t > 0$

$2 {t}^{2} - 5t + 3 = 0$

$a = 2$

$b = - 5$

$c = 3$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 5)}^{2} - 4 \times 2 \times 3$

$\Delta = 25 - 24$

$\Delta = 1$

$t_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 5) \pm \sqrt{1} }{2 \times 2} = \frac{5 \pm1}{4}$

$t_{1}= \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2} = > {( \frac{9}{4}) }^{x} = \frac{3}{2}$

${( \frac{ {3}^{2} }{ {2}^{2} } )}^{x} = \frac{3}{2} = > {[{(\frac{3}{2})}^{2}]}^{x} = \frac{3}{2} = > {( \frac{3}{2}) }^{2x} = {( \frac{3}{2} )}^{1} = > 2x = 1 = > x_{1} = \frac{1}{2}$

$t_{2} = \frac{5 - 1}{4} = \frac{4}{4} = 1 = > {( \frac{9}{4}) }^{x} = 1 = > x_{2} = 0$

$e)lg( {10}^{2x + 1} + 7 \times {10}^{x + 1}) = lg( {10}^{x + 2} - 20)$

Condiții de existență :

${10}^{2x + 1} + 7 \times {10}^{x + 1} > 0$

${10}^{x + 2} - 20 > 0$

${10}^{2x + 1} + 7 \times {10}^{x + 1} = {10}^{x + 2} - 20$

${10}^{2x + 1} + 7 \times {10}^{x + 1} - {10}^{x + 2} + 20 = 0 \: | \div 10$

${10}^{2x} + 7 \times {10}^{x} - {10}^{x + 1} + 2 = 0$

${ ({10}^{x} )}^{2} + 7 \times {10}^{x} - {10}^{x} \times 10 + 2 = 0$

${10}^{x} = t \: \: \: ,t > 0$

${t}^{2} + 7t - 10t + 2 = 0$

${t}^{2} - 3t + 2 = 0$

${t}^{2} - 2t - t + 2 = 0$

$t(t - 2) - (t - 2) = 0$

$(t - 2)(t - 1) = 0$

$t - 2 = 0 = > t_{1} = 2 = > {10}^{x} = 2$

${10}^{x} = 2 = > x_{1} = log_{10}(2)$

$t - 1 = 0 = > t_{2} = 1 = > {10}^{x} = 1 = > x_{2} = 0$

$f) {lg}^{2} x + 4lgx + 3 = 0$

Condiția de existență :

$x > 0$

$lgx = t$

${t}^{2} + 4t + 3 = 0$

${t}^{2} + 3t + t + 3 = 0$

$t(t + 3) + t + 3 = 0$

$(t + 3)(t + 1) = 0$

$t + 3 = 0 = > t_{1} = - 3 = > lgx = - 3 = > x_{1} = {10}^{ - 3} = \frac{1}{ {10}^{3} } = \frac{1}{1000}$

$t + 1 = 0 = > t_{2} = - 1 = > lgx = - 1 = > x_{2} = {10}^{ - 1} = \frac{1}{10}$

$g) log_{2}(3x - 1) = log_{2}( \frac{1}{2} )$

Condiția de existență :

$3x - 1 > 0$

$3x - 1 = \frac{1}{2} \: | \times 2$

$6x - 2 = 1$

$6x = 1 + 2$

$6x = 3 \: | \div 6$

$x = \frac{3}{6}$

$x = \frac{1}{2}$