Умоляю помогите!!! Решить показатель не уравнения: 7^x+10=49 3-4^x+5=2 3^x+3^x+1=4...

${7}^{x + 10} = 49 \\ {7}^{x + 10} = {7}^{2} \\ x + 10 = 2 \\ x = - 8 \\ \\$
$3-4^{x+5}=2 \\ 4^{x+5} = 1 \\ 4^{x+5} = {4}^{0} \\ x + 5 = 0 \\ x = - 5 \\ \\$
$3^x+3^{x+1}=4 \\ 3^x(1 + 3) = 4 \\ {3}^{x} = 1 \\ 3^x = {3}^{0} \\ x = 0 \\ \\$

0 \\ {y}^{2} - 3y - 4 = 0 \\ (y - 4)(y + 1) = 0 \\ y_1 = 4 \\ y_2= - 1 \\ {2}^{x} = 4 \\ x = 2 \\ " alt="4^x-3 \cdot 2^x=4 \\ ( {2}^{x} ) ^{2} - 3 \cdot 2^x - 4 = 0\\ y = {2}^{x} > 0 \\ {y}^{2} - 3y - 4 = 0 \\ (y - 4)(y + 1) = 0 \\ y_1 = 4 \\ y_2= - 1 \\ {2}^{x} = 4 \\ x = 2 \\ " align="absmiddle" class="latex-formula">
у2 нам не подходит, так как

0" alt="y = {2}^{x} > 0" align="absmiddle" class="latex-formula">