Очень срочно! Помогите решить задания по алгебре 9 класс. 

1) [latex]26^{26-x}=4[/latex] [latex]26-x=log_{26}(4)[/latex] [latex]x=26-log_{26}(4)[/latex] 2) [latex]8=4^{ \frac{1}{26x+1} }[/latex] [latex] \frac{1}{26x+1}=log_4(8) [/latex] [latex]26x+1= \frac{1}{log_4(8)} =log_8(4)[/latex] [latex]x= \frac{log_8(4)-1}{26} [/latex] 3) [latex]( \frac{12}{41} )^{ \frac{x}{26} +1}=( \frac{51}{41} )^{ \frac{x}{26} +1}[/latex] [latex](\frac{12}{41}:\frac{51}{41})^{ \frac{x}{26} +1}=1[/latex] [latex](\frac{12}{51})^{\frac{x}{26} +1}=1[/latex] [latex]\frac{x}{26} +1=0[/latex] [latex]x=- \frac{1}{26} [/latex] 4) [latex]14^{26x}-14^{26x-1}=13[/latex] [latex]14^{26x}- \frac{1}{14}*14^{26x}=14^{26x}* \frac{13}{14} =13 [/latex] [latex]14^{26x}* \frac{1}{14} =14^{26x-1}=1[/latex] [latex]26x-1=0[/latex] [latex]x= \frac{1}{26} [/latex] 5) [latex]( \frac{7}{8} )^{ \frac{x-1}{2} }= \sqrt[26]{\frac{8}{7}}=( \frac{8}{7} )^{ \frac{1}{26} } =( \frac{7}{8} )^{- \frac{1}{26} }[/latex] [latex] \frac{x-1}{2}=- \frac{1}{26} [/latex] [latex]x-1=- \frac{2}{26} =- \frac{1}{13} [/latex] [latex]x = 1 - \frac{1}{13} = \frac{12}{13} [/latex] 6) [latex]2^x+2^{x-3}=18[/latex] [latex]2^x+ \frac{1}{8}*2^x=2^x*(1+ \frac{1}{8} ) =2^x* \frac{9}{8} =18[/latex] [latex]2^x* \frac{1}{8} =2^{x-3}=2[/latex] x - 3 = 1 x = 4 7) [latex]4^x - 6*2^x+8=0[/latex] Замена 2^x = y > 0 при любом х y^2 - 6y + 8 = 0 (y - 2)(y - 4) = 0 y1 = 2^x = 2; x1 = 1 y2 = 2^x = 4; x2 = 2