Question

Solve : $4\cdot 2^{2x}-6^x={18.3}^{2x}$

Answer 1

${4.2}^{2x}-6^x={18.3}^{2x}$

$4.(2^x{)}^2-2^x{.3}^x=18(3^x{)}^2$

$4P^2-PQ-18Q^2=0$ Let $2^x=P,3^x=Q$

$4P^2+8PQ-9PQ-18Q^2=0$

$4P(P+2Q)-9Q(P+2Q)=0$

$(P+2Q)(4P-9Q)=0$

case 1 : - When 4P-9Q = 0

${4.2}^x-{9.3}^x=0$

$2^{x+2}-3^{x+2}=0$

$(\frac{2}{3}{)}^{x+2}=1\Rightarrow x+2=0\Rightarrow x=-2.$

case 2 :- when P+2Q = 0

P = -2Q.

$=2^x={2.3}^x$ & we know exponential function can't be negative.

$=2^x\ne -{2.3}^x$

the ans is x = -2