Question

√6^x-9^x= 2^x-3^x
Va rog ajutati-ma, e urgent!

Answer 1

$6^{x}-9^x=(2^x-3^x)^2 \\ 6^x-9^x=4^x-2*6^x+9^x \\ 4^x-3*6^x+2*9^x=0 \\ 1-3*( \frac{3}{2})^x+2*( \frac{3}{2}) ^{2x} =0 \\ (\frac{3}{2})^x =t,t\ \textgreater \ 0 \\ 2t^2-3t+1=0 \\ t= 1;t= \frac{1}{2}$
$( \frac{3}{2})^x=1;x=0 \\ ( \frac{3}{2})^x= \frac{1}{2};x= log_{ \frac{3}{2} } \frac{1}{2}$