Помогите пожалуйста с решением[tex]\sqrt{3^{x}} =27^{2/3}[/tex] (sin п/3)^2х-2 =2/4. Created by CHudesnayaLedi. algebra-ru

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CHudesnayaLedi @CHudesnayaLedi

July 2022 1 11 Report

Помогите пожалуйста с решением

[tex]\sqrt{3^{x}} =27^{2/3}[/tex]

(sin п/3)^2х-2 =2/4

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NNNLLL54

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$1)\; \; \sqrt{3^{x}}=27^{2/3}\; \; \Rightarrow \; \; 3^{x/2}=(3^3)^{2/3}\; ,\; \; \; 3^{x/2}=3^2\\\\\frac{x}{2}=2\; \ \Rightarrow \; \; \; \underline {x=4}\\\\\\2)\; \; (sin\frac{\pi}{3})^{2x-2}=\frac{2}{4}\; \; \Rightarrow \; \; \; (\frac{\sqrt3}{2})^{2x-2}=\frac{1}{2}\; ,\\\\2x-2=log_{\sqrt3/2}\, (\frac{1}{2})\; \; ,\; \; 2x-2=-log_{\sqrt3/2}\, 2\; \; ,\; \; 2x-2=-\frac{1}{log_2\frac{\sqrt3}{2}}\; ,\\\\2x-2=-\frac{1}{log_2\sqrt3-log_22}\; \; ,\; \; 2x-2=-\frac{1}{\frac{1}{2}log_23-1}\; \; ,\; \; 2x-2=-\frac{2}{log_23-2}\; ,$

$x-1=-\frac{1}{log_23-2}\; \; ,\; \; x=1-\frac{1}{log_23-2}\; \; ,\; \; x=\frac{log_23-3}{log_23-2}$

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