Exponenciální rovnice

Exponenciální rovnice obsahují neznámou v exponentu. Rovnice řešíme tak, že se snažíme převést všechny členy na stejný základ

Poté už jen porovnáváme exponenty a řešíme vzniklou rovnici (většinou lineární nebo kvadratickou). V některých případech užíváme substituci.

Někdy převod na stejné základy nelze provést a na každé straně vznikne člen s jiným základem, potom musíme rovnici logaritmovat.


Při úpravách užíváme vztahy pro práci s exponenty:

Umocňování součinu




Násobení čísel o stejném základu




Umocňování




Dělení čísel o stejném základu




Převod odmocniny na exponent




Převody mezi exponenciálním tvarem a zlomkovým tvarem čísla








$\huge 2^{3x-1} . 4 = 8^{x+1}.(\frac{1}{2})^{x}$

$\large 2^{3x} .2^{-1}.2^{2} = 2^{3x}.2^{3}.2^{-x}$

$\large 2^{3x-1+2} = 2^{3x+3-x}$

$\large 2^{3x+1} = 2^{2x+3}$

$\large 3x+1 = 2x+3$

$\large 3x-2x = 3-1$

$\large x = 2$

$\huge \sqrt[4]{4^{x}} . \sqrt[3]{2^{x-3}}= \sqrt[6]{16}$

$\large \sqrt[4]{2^{2x}} . \sqrt[3]{2^{x-3}}= \sqrt[6]{2^{4}}$

$\large 2^{\frac{2x}{4}} . 2^{\frac{x-3}{3}}= 2^{\frac{4}{6}}$

$\large \frac{2x}{4} + \frac{x-3}{3}= \frac{4}{6}$

$\large \frac{x}{2} + \frac{x-3}{3}= \frac{2}{3} \;/\times 6$

$\large 3x + 2x-6= 4$

$\large 5x= 10$

$\large x=2$

$\huge \frac{1}{3^{x}}= \frac {1}{\sqrt{3}} . \sqrt[6]{27^{3-3x}}.{(\frac{1}{9})}^{x+3}$

$\large 3^{-x}= 3^{-\frac{1}{2}} . 3^{\frac{3(3-3x)}{6}}. 3^{-2(x+3)}$

$\large -x= -\frac{1}{2}+\frac{3(3-3x)}{6}-2(x+3)$

$\large -x= -\frac{1}{2}+\frac{(3-3x)}{2}-2(x+3)\;/\times 2$

$\large -2x= -1+3-3x-4(x+3)$

$\large -2x= 2-3x-4x-12$

$\large 5x= -10$

$\large x= -2$

$\huge 0,25 . {(\frac{1}{4})}^{2x} = 1$

$\large (\frac{1}{4})^{1} . (\frac{1}{4})^{2x} = (\frac{1}{4})^{0}$

$\large 1+2x = 0$

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

$\huge 2 . {0,5}^{x^{2}+\frac{8}{3}x} = \frac {8}{\sqrt[3]{4}}$

$\large 2^{1} . {(\frac{1}{2})}^{x^{2}+\frac{8}{3}x} = \frac {2^{3}}{\sqrt[3]{2^{2}}}$

$\large 2^{1} . {2}^{-x^{2}-\frac{8}{3}x} = 2^{3} . 2^{-\frac{2}{3}}$

$\large 1- x^{2} -\frac{8}{3}x = 3 - \frac{2}{3}$

$\large - x^{2} -\frac{8}{3}x - \frac{4}{3} = 0 \;/\times -3$

$\large 3x^{2} + 8x + 4 = 0$

$\large x_{1,2} = \frac{-8\pm \sqrt{64-48}}{6}$

$\large x_{1,2} = \frac{-8\pm 4}{6}$

$\large x_{1} =  -\frac{2}{3}$

$\large x_{2} = -2$

$\huge 5^{x} . 2^{x} = 100^{x-1}$

$\large 10^{x} = 10^{2(x-1)}$

$\large x = 2x-2$

$\large x =2$

$\huge \frac{81}{16} = (\frac{2}{3})^{x} . (\frac{9}{4})^{x+1}$

$\large (\frac{3}{2})^{4} = (\frac{2}{3})^{x} . (\frac{3}{2})^{2(x+1)}$

$\large (\frac{3}{2})^{4} = (\frac{3}{2})^{-x} . (\frac{3}{2})^{2x+2)}$

$\large 4 = - x + 2x + 2$

$\large x = 2$

$\huge \frac{3^{x}}{2 . 3^{\sqrt{3}}} = 4,5$

$\large \frac{3^{x}}{2 . 3^{\sqrt{3}}} = \frac{9}{2} \;/\times 2 \times 3^{\sqrt{3}} $

$\large 3^{x} = 3^{2} . 3^{\sqrt{3}}$

$\large x = 2 + \sqrt{3}$

$\huge 3^{x} + 3^{x+1} = 108$

$\large 3^{x} + 3^{x} . 3^{1} = 108$

$\large 3^{x}( 1 +. 3^{1}) = 108$

$\large 3^{x} . 4= 108$

$\large 3^{x} = \frac {108}{4}$

$\large 3^{x} = 27$

$\large x = 3$

$\huge 2^{x+1} + 2^{x-1} + 2^{x+3}= \frac {21}{8}$

$\large 2^{x} . 2^{1} + 2^{x} . 2^{-1} + 2^{x} . 2^{3}= \frac {21}{8}$

$\large 2^{x}.( 2^{1} + 2^{-1} +  2^{3}) = \frac {21}{8}$

$\large 2^{x}.(\frac{4+1+16}{2}) = \frac {21}{8}$

$\large 2^{x}.\frac {21}{2} = \frac {21}{8} \;/\times \frac {2}{21}$

$\large 2^{x} = \frac {21}{8} \times \frac {2}{21}$

$\large 2^{x} = \frac {1}{4}$

$\large 2^{x} = 2^{-2}$

$\large x = -2$

$\huge 7 . 4^{-x+2} = 3 . 4^{-x+3} - 5$

$\large 7 . 4^{-x} . 4^{+2} - 3 . 4^{-x} . 4^{3} = - 5$

$\large 4^{-x}.(7 . . 4^{+2} - 3 .  4^{3}) = - 5$

$\large 4^{-x}.(112-192) = - 5$

$\large 4^{-x}.(-80 )= - 5 \;/\div -80$

$\large 4^{-x} = \frac{5}{80}$

$\large 4^{-x} = \frac{1}{16}$

$\large 4^{x} = 16$

$\large x = 2$

$\huge 3^{x} . (\frac{1}{2})^{x} + 3^{x+1} . (\frac{1}{2})^{x+1}= \frac{5}{3}$

$\large 3^{x} . (\frac{1}{2})^{x} + 3^{x} . 3^{1} . (\frac{1}{2})^{x} . (\frac{1}{2})^{1} = \frac{5}{3}$

$\large (\frac{3}{2})^{x} + (\frac{3}{2})^{x} . (\frac{3}{2})^{1} = \frac{5}{3}$

$\large (\frac{3}{2})^{x} . (1  + \frac{3}{2}) = \frac{5}{3}$

$\large (\frac{3}{2})^{x} . \frac{5}{2} = \frac{5}{3} \;/\times \frac {2}{5}$

$\large (\frac{3}{2})^{x}  = \frac{5}{3} . \frac {2}{5}$

$\large (\frac{3}{2})^{x}  = \frac{2}{3}$

$\large x=-1$

$\huge \frac {1}{4} . 2^{x} + \frac{1}{2} . 4^{x} =9$

$\large \frac {2^{x}}{4} + \frac{2^{2x}}{2} =9\;/\times 4$

$\large 2^{x} + 2 . 2^{2x} =36$

$\large 2 . 2^{2x} + 2^{x} - 36 = 0$

$\large S: 2^{x} = y$

$\large 2y^{2} + y - 36 = 0$

$\large y_{1,2} = \frac{-1\pm \sqrt{1+288}}{4}$

$\large y_{1,2} = \frac{-1\pm 17}{4}$

$\large y_{1} =  4$

$\large y_{2} = -\frac{9}{2}$

$\large S(y_{1}): 2^{x} = 4$

$\large x=2$

$\large S(y_{2}): 2^{x} \neq -\frac{9}{2}$

$\huge 3^{x} + 3^{x+1}= 7 . 4^{x} - 4^{x+1}$

$\large 3^{x} + 3^{x} . 3^{1}= 7 . 4^{x} - 4^{x} . 4^{1}$

$\large 3^{x}(1 +  3^{1}) = 4^{x}(7  -  4^{1})$

$\large 3^{x} . 4 = 4^{x} . 3$

$\large \frac{3^{x}}{4^{x}} = \frac {3}{4}$

$\large (\frac{3}{4})^{x}= (\frac{3}{4})^{1}$

$\large x = 1$

$\huge 2^{x-1} -2^{x-2}= 5^{x-3} + 2^{x-3}$

$\large 2^{x} . 2^{-1} -2^{x} . 2^{-2} = 5^{x} . 5^{-3} + 2^{x} . 2^{-3}$

$\large 2^{x} . 2^{-1} -2^{x} . 2^{-2} - 2^{x} . 2^{-3}= 5^{x} . 5^{-3}$

$\large 2^{x}(2^{-1} - 2^{-2} - 2^{-3})= 5^{x} . 5^{-3}$

$\large 2^{x} . (\frac{4-2-1}{8})= 5^{x} . 5^{-3}$

$\large 2^{x} . 2^{-3}= 5^{x} . 5^{-3}$

$\large \frac{2^{x}}{5^{x}} = \frac { 5^{-3}}{ 2^{-3}}$

$\large (\frac{2}{5})^{x} =(\frac{2}{5})^{3}$

$\large x=3$

$\huge 2 . 4^{x} + 5^{x-\frac{1}{2}}= 5^{x+\frac{1}{2}} - 2^{2x-1}$

$\large 2 . 2^{2x} + 2^{2x}.2^{-1}= 5^{x}.5^{\frac{1}{2}} - 5^{x}.5^{-\frac{1}{2}}$

$\large 2 . 4^{x} + 4^{x}.2^{-1}= 5^{x}.5^{\frac{1}{2}} - 5^{x}.5^{-\frac{1}{2}}$

$\large 4^{x}(2  + 2^{-1})= 5^{x}.(5^{\frac{1}{2}} - 5^{-\frac{1}{2}})$

$\large 4^{x}(4^{\frac{1}{2}} + 4^{-\frac{1}{2}})= 5^{x}.(5^{\frac{1}{2}} - 5^{-\frac{1}{2}})$

$\large \frac{4^{x}}{5^{x}}= .\frac{5^{\frac{1}{2}} - 5^{-\frac{1}{2}}}{4^{\frac{1}{2}} + 4^{-\frac{1}{2}}}$

$\large (\frac{4}{5})^{x} = .\frac{5^{\frac{1}{2}} - 5^{-\frac{1}{2}}}{4^{\frac{1}{2}} + 4^{-\frac{1}{2}}}$

$\large (\frac{4}{5})^{x} = \frac{\sqrt{5} - \frac{1}{\sqrt{5}}}{\sqrt{4} + \frac{1}{\sqrt{4}}}$

$\large (\frac{4}{5})^{x} = \frac{\frac{5-1}{\sqrt{5}}}{\frac{4+1}{\sqrt{4}}}$

$\large (\frac{4}{5})^{x} = \frac{\frac{4}{\sqrt{5}}}{\frac{5}{\sqrt{4}}}$

$\large (\frac{4}{5})^{x} = \frac{4}{\sqrt{5}} . \frac{\sqrt{4}}{5}$

$\large (\frac{4}{5})^{x} = \frac{4^{1} . 4^\frac{1}{2}}{5^{1} . 5^\frac{1}{2}}$

$\large (\frac{4}{5})^{x} = \frac{4^\frac{3}{2}}{5^\frac{3}{2}}$

$\large (\frac{4}{5})^{x} = (\frac{4}{5})^{\frac{3}{2}}$

$\large x = \frac{3}{2}$

$\huge 3^{x} + \frac{9^{x}}{3}= 3^{x+1} + \frac{9^{x}}{9}$

$\large 3^{x} + \frac{3^{2x}}{3}= 3^{x}.3^{1} + \frac{3^{2x}}{9}$

$\large S: 3^{x} = y$

$\large y + \frac{y^{2}}{3}= 3y + \frac{y^{2}}{9} \;/\times 9$

$\large 9y + 3y^{2}= 27y + y^{2}$

$\large 2y^{2} - 18y = 0$

$\large 2y(y-9) = 0$

$\large y_{1} =  0$

$\large y_{2} =9$

$\large S: 3^{x} =9$

$\large x = 2$

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