Помогите решить. Тема: Степень с действительным показателем.
[tex]9^{3x+4} \sqrt{3}= \frac{ 27^{x-1} }{ \sqrt{3}} [/tex]
[tex] ( \sqrt[3]{2})^{x-1} =( \frac{2}{ \sqrt[3]{2}}) ^{2x} [/tex]
[tex]9^{3x+4} \sqrt{3}= \frac{ 27^{x-1} }{ \sqrt{3}} [/tex]
[tex] ( \sqrt[3]{2})^{x-1} =( \frac{2}{ \sqrt[3]{2}}) ^{2x} [/tex]
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√3 = 3^(1/2)9^(3x+4) = (3^2)^(3x+4) = 3^(6x+8)
получилось:
3^(6x+8+1/2) = 3^(3x-3-1/2)
6x+8.5 = 3x-3.5
3x = -12
x = -4
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2^((x-1)/3) = 2^((1-1/3)*2x)
(x-1)/3 = 2x*2/3
x-1 = 4x
3x = -1
x = -1/3