Question

Solve the two questions given in picture above write in a book and upload them

Answer 1

Answer:

For both questions, remember that when bases are smae, powers are equal.

$19) i)2^{-3} =\frac{1}{2^{x}} \\\\=>2^{-3} =2^{-x} \\=>-3=-x\\=>x=3$

$ii)\frac{1}{49} *7^{2x} =7^{8} \\\\=>\frac{1}{7^{2} } *7^{2x} =7^{8}\\\\=>7^{-2}  *7^{2x} =7^{8}\\=>7^{-2+2x}=7^{8} \\=>-2+2x=8\\=>2x=8+2\\=>2x=10\\=>x=10/2\\=>x=5$

$iii)(3^{6} )^{4} =3^{12x} \\=>3^{6*4} =3^{12x}\\=>3^{24} =3^{12x}\\=>24=12x\\=>x=24/12\\=>x=2$

$20)  i)(-4)^{n+1} *(-4)^{-7} =(-4)^{16} \\=>(-4)^{n+1+(-7)}=(-4)^{16}\\=>(-4)^{n-6}=(-4)^{16}\\=>n-6=16\\=>n=16+6\\=>n=22$

$ii)7^{2n+1}$÷$49=7^{3}$

$=>7^{2n+1}$÷$7^{2} =7^{3}$

$=> 7^{2n+1-2}=7^{3}  \\=> 7^{2n-1}=7^{3}\\=> 2n-1=3\\=>2n=3+1\\=>2n=4\\=>n=4/2\\=>n=2$

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