Solve system of equations: $7^{\log_3\sqrt{x}}-3^{\log_{81}y}=4 \land 7^{\log_3x}-3^{\log_9y}=40$

Question

$$7^{\log_3\sqrt{x}}-3^{\log_{81}y}=4 \land 7^{\log_3x}-3^{\log_9y}=40$$ I cant finish any of the two equations. On the first I get to this part $7^{\frac{1}{2}\log_3x}-3^{\frac{1}{4}\log_3y}=4$ and in the second I get something similar like in this one. What do I do next?

Answer 1

Let $A=7^{\frac12\log_3 x}$, and let $B=3^{\log_{81}y}=3^{\frac12\log_9 y}$

Then your two equations become: $$A-B=4$$ and $$A^2-B^2=40$$

Does that help?

By the way, $B$ simlifies to $B=y^{1/4}$.

Answer 2

Hint: write $$7^{\frac{\ln(x)}{2\ln(3)}}-3^{\frac{\ln(y)}{4\ln(3)}}=4$$ and $$7^{\frac{\ln(x)}{\ln(3)}}-3^{\frac{\ln(y)}{2\ln(3)}}=40$$ And then you Can write$$A^{1/2}+B^{1/4}=4$$ $$A-B^{1/2}=40$$ And then we get $$A=16+B^{1/2}-8B^{1/4}$$ and $$A=40+B^{1/2}$$