Question

please solve the problem
please​

Answer 1

Answer:

a) $\frac{1}{8}$

b)$\frac{1}{1728}$

Step-by-step explanation:

$5^{x} =125^{-1}\\=>5^{x} ={(5^{3})}^{-1} \\=>5^{x} =5^{-3}\\Since,\ bases \ are \ same\ \\Therefore,\ equating\ the \ powers\\x=-3$

a)

$2^{x}\\=>2^{-3}\\\\=>\frac{1}{2^{3} }\\\\=>\frac{1}{8}$

b)

$3^{x}\times 4^{x}\\=>(3^{-3}) (4^{-3} ) \\=>{(3\times 4) }^{-3} \\=>12^{-3} \\\\=>\frac{1}{12^{3} } \\\\=>\frac{1}{1728}$

Answer 2

Answer:

Given,

$\: \: \: \: \: {5}^{x} = {125}^{ - 1} \\ \implies \: {5}^{x} =( { {5}^{3} })^{ - 1} \\ \implies \: {5}^{x} = {5}^{ - 3} \\ \therefore \: \: x = - 3$

$\sf \: cause \: \\ \: if \: \: \: \: \: \: \: \: {a}^{m} = {a}^{n} \: \: \: \: \: then \: \: \: \: \: \: m = n \:$

Now,

a)

${2}^{x} \\ = {2}^{ - 3} \\ =( { {2}^{3} })^{ - 1} \\ = {8}^{ - 1} \\ = \frac{1}{8}$

b)

${3}^{x} \times {4}^{x} \\ = {3}^{ - 3} \times {4}^{ - 3} \\ = ( { {3}^{3} })^{ - 1} \times ( { {4}^{3} })^{ - 1} \\ = {27}^{ - 1} \times {64}^{ - 1} \\ = \frac{1}{27} \times \frac{1}{64} \\ = \frac{1}{1728}$