Question

Determinati in fiecare dintre urm cazuri nr real x pentru care egalitatea respectiva este adevarata :
a)log din 2 la x +log din 4 la x= 6
b)log din 3 la x + log din 9 la x= 3
c)log din 4 la x - log din 2 la x = 0

Answer 1

[tex]a)\log_2x +\log_4x= 6\\ C.E.\:x\ \textgreater \ 0\\ \log_2x +\log_{2^2}x= 6\\ \log_2x +\frac{1}{2}\log_{2}x= 6\\ \log_2x +\log_{2}x^{\frac{1}{2}}= 6\\ \log_2{x\cdot x^{\frac{1}{2}}}=6\\ x\cdot x^{\frac{1}{2}}=2^6\\ x^{1+\frac{1}{2}=2^6\\ x^{\frac{3}{2}=2^6\\ \sqrt{x^3}=2^6\\ x^3=(2^6)^2=2^{12}\\ x=\sqrt[3]{2^{12}}=2^4=16[/tex]

[tex]b)\log_3x + \log_9x= 3\\ C.E.\:x >[tex]c)\log_4x - \log_2x = 0\\ C.E.\:x \ \textgreater \ 0\\ \log_{2^2}x=\log_2x\\ \frac{1}{2}\log_2x=\log_2x\\ \log_2x^{\frac{1}{2}}=\log_2x\\ x^{\frac{1}{2}}=x\\ \sqrt{x}=x\\x=x^2\\x^2-x=0\Leftrightarrow x(x-1)=0\Leftrightarrow(x=0\:sau\:x=1)\\ dar\:x\ \textgreater \ 0(\:din\:C.E.)\Leftrightarrow x=1[/tex] 0\\ \log_3x + \log_{3^2}x= 3\\ \log_3x + \frac{1}{2}\log_3x= 3\\ \log_3x + \log_3x^{ \frac{1}{2}}= 3\\ \log_3{(x\cdot x^{ \frac{1}{2}})}=3\\ x^\frac{3}{2}=3^3\\ \sqrt{x^3}=3^3\\ x^3=(3^3)^2\\ x=\sqrt[3]{3^6}=3^2=9[/tex]