1) Solve 7 to the power of x+1 divided by 49 to the power of y = 343, and 81 to the power of y times (1/27) to the power of x = 9 simultaneously.
2) Solve the equation 6 to the power of 2x-1 x 36 to the power of x = 1
I'm thinking that it might be helpful if you put brackets around certain parts in order to make the equation not ambiguous. Eg if a = 81, and b = y/27, and c = x, then you're looking at a^(b^c) = 9 ?
And if a = 7, b = x+1, c = 49, d = y, then are you looking at (a^b)/(c^d)? Or a^{(b/c)^d} etc ?
And also same with the other equation..... if a = 6, and b = 36*(2x-1), and c = x, then you're looking at a^(b^c) = 1 ?
$\displaystyle \fbox{\dfrac{7^{x+1}}{49^y}=343} $ $\displaystyle \implies \dfrac{7^x\times 7^1}{49^y}=343 \implies \frac{7^x}{49^y}=\frac{343}{7^1}\implies \frac{7^x}{49^y}=\frac{49}{1}$ $\displaystyle \implies 7^x = 49^{y+1} \implies 7^x = \left(7^2\right)^{y+1} \implies 7^x = 7^{2y+2} \implies x = 2y+2$
$\displaystyle \fbox{81^{y}\left(\dfrac{1}{27}\right)^x=9}$ $\displaystyle \implies \left(9^2\right)^{y}\left(\tfrac{1}{27}\right)^{2y +2}=9$ $\displaystyle \implies 9^{2y}\left(\tfrac{1}{27}\right)^{2y} \times \left(\tfrac{1}{27}\right)^2=9 \implies \left(\tfrac{9^{2y}}{27^{2y}}\right) \times \left(\tfrac{1}{27}\right)^2 = 9$ $\displaystyle \implies \left(\tfrac{1}{3}\right)^{2y} \times \left(\tfrac{1}{9}\times\tfrac{1}{3}\right)^2 = 9 $ $\displaystyle \implies \left(\tfrac{1}{9}\right)^y \times \left(\tfrac{1}{9}\right)^2 \times \left(\tfrac{1}{3}\right)^2 =9 \implies \left(\tfrac{1}{9}\right)^y \times \left(\tfrac{1}{9}\right)^3 = 9$
$\displaystyle \implies \left(\tfrac{1}{9}\right)^y = 9^4 \implies \left(\tfrac{1}{9}\right)^{y}=(\tfrac{1}{9})^{-4} \implies y=-4$
$\displaystyle y=-4 \implies x=2(-4)+2=-6$