Question

$ 3^{x} =10 , 2^{y} =5 , ( 4^{y+1} -19)^{x+1} =?$

Answer 1

$ 3^{x}=10 \rightarrow 3^{1}. 3^{x} =3.10 \rightarrow 3^{x+1} =30 \rightarrow lox_{3} 30=x+1$

$ 2^{y} =5 \rightarrow 2^{1} 2^{y} =2.5 \rightarrow 2^{y+1} =10 \rightarrow log_{2}10 =y+1$

$$ ( 4^{y+1} -19)^{x+1} $$

$$ ( 4^{ log_{2} 10} -19)^{x+1} $$

$$ ( 10^{ log_{2} 4} -19)^{x+1} \rightarrow 81^{x+1} $$

$$ 81^{ log_{3}30 } $$

$$ 30^{ log_{3}81 } $$

$$ 30^{4} $$

Answer 2

$$(4(2^y)^2-19)^{x+1}=(4(25)-19)^{x+1}=81^{x+1}=81(3^x)^4=30^4$$