Question

sa s erezolve ecuatia ( exponentiala )

b) $49^{x} - 35^{x} = 25^{x}$

Answer 1

[tex] 49^{x} - 35^{x} = 25^{x} ; 7^{2x} - 5^{x} X 7^{x} = 5^{2x} | : 5^{2x} \\ \frac{ 7^{2x} }{ 5^{2x} } - \frac{ 5^{x} X 7^{x} }{ 5^{2x} } = 1 \\ (\frac{7}{5}) ^{2x} - ( \frac{7}{5} )^{x} - 1= 0 \\ ( \frac{7}{5} )^{x} = t ; t\ \textgreater \ 0 \\ t^{2} - t - 1 =0 \\ t_{12} = \frac{1 +- \sqrt{5} }{2} \\ (\frac{7}{5}) ^{x} = \frac{1 + \sqrt{5} }{2} | lg \\ lg_{10} ( \frac{7}{5}) ^{x} = lg_{10} ( \frac{1 + \sqrt{5} }{2} ) \\ x lg_{10} ( \frac{7}{5} ) = lg_{10} [/tex] din (1+ radical5)/2
scoti x-ul din ecuatie in functie de ce ai adica logaritmii respectivi