Question

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Answer 1

$x^4+3^{3*\frac{1}{3} log_{3^3}^{(x^2+1)}-2^{4log_{2}^x}}$

$x^4+3^{3log_{3^3}^{(x^2+1)}-2^{log_{2^{\frac{1}{4}} ^x}}$1. C

2. $\sqrt[9]{(-2)^{15}} =\sqrt[9]{(-2^5)^{3}}= \sqrt[3]{(-2)^{5}}$ deci A

3. C

4.  afli pe rand valoarea fiecarui logaritm

$36^{log_6^{5}}=(6^2)^{log_6^{5}}=6^{2log_6^{5}}=6^{log_6^{5^2}}=5^2 = 25$

$10^{1-lg2}=10^1 *10^{-lg2} = 10*2^{-1}=10*\frac{1}{2} =5$

$3^{log_9^{36}} = 3^{log_{3^2}^{36}=3^{log_3^{36^{\frac{1}{2}}}}=36^{\frac{1}{2}}={6^2}^{\frac{1}{2}} = 6$

⇒ 25+5-6 = 24

5. nu ma descurc

6. a) $\sqrt[6]{(1-\sqrt{3} )^2}=\sqrt[3]{(1-\sqrt{3}) }=(1-\sqrt{3} )^{\frac{1}{3}$

b) $x^4+3^{log_{27}^{(x^2+1)}-2^{log_{\sqrt[4]{2}^x}}$

$x^4+3^{3log_{3^3}^{(x^2+1)}-2^{log_{2^{\frac{1}{4} }^x}}$

$x^4+3^{3*\frac{1}{3} log_{3}^{(x^2+1)}-2^{4log_{2^x}}$

$x^4+x^2+1-x^4 = x^2+1$