If 3x−y=12, what is the value of [tex] \frac{ {8}^{x} }{ {2}^{y} }[/tex] A) [tex] {2}^{12} [/tex] B) [tex] {4}^{4} [/tex] C) [tex] {8}^{2} [/tex] D) The value cannot be determined from the information given. About the author Madelyn
Answer: option (A) 2¹² Step-by-step explanation: Given : 3x – y = 12 To find : the value of \sf \dfrac{8^x}{2^y} 2 y 8 x Solution : 3x – y = 12 3x = 12 + y —[1] Now, 8 can be written as (2 × 2 × 2) i.e., 8 = 2³ \longrightarrow \sf 8^x = (2^3)^x = 2^{3x}⟶8 x =(2 3 ) x =2 3x Put 3x = 12 + y [ ∵ eqn. 1 ] \longrightarrow \sf 2^{3x}=2^{12+y}⟶2 3x =2 12+y So, \begin{gathered}\implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{3x}}{2^y} \\\\ \implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{12+y}}{2^y}\end{gathered} ⟹ 2 y 8 x = 2 y 2 3x ⟹ 2 y 8 x = 2 y 2 12+y We know, \rm \dfrac{a^m}{a^n}=a^{m-n} a n a m =a m−n \begin{gathered}\implies \sf \dfrac{8^x}{2^y}=2^{(12+y-y)} \\\\ \implies \sf \dfrac{8^x}{2^y}=2^{12}\end{gathered} ⟹ 2 y 8 x =2 (12+y−y) ⟹ 2 y 8 x =2 12 Therefore, the required value is 2¹² ________________________________ Laws of exponents : \begin{gathered}\dag \ \sf a^m \times a^n=a^{m+n} \\\\ \dag \ \sf \dfrac{a^m}{a^n}=a^{m-n} \\\\ \dag \ \sf (a^m)^n=a^{mn} \\\\ \dag \ \sf a^{-n}=\dfrac{1}{a^n} \\\\ \dag \ \sf \sqrt[n]{a} =a^{\dfrac{1}{n}\end{gathered} Step-by-step explanation: pls mark brainliest Reply
Answer: option (A) 2¹² Step-by-step explanation: Given : 3x – y = 12 To find : the value of [tex]\sf \dfrac{8^x}{2^y}[/tex] Solution : 3x – y = 12 3x = 12 + y —[1] Now, 8 can be written as (2 × 2 × 2) i.e., 8 = 2³ [tex]\longrightarrow \sf 8^x = (2^3)^x = 2^{3x}[/tex] Put 3x = 12 + y [ ∵ eqn. 1 ] [tex]\longrightarrow \sf 2^{3x}=2^{12+y}[/tex] So, [tex]\implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{3x}}{2^y} \\\\ \implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{12+y}}{2^y}[/tex] We know, [tex]\rm \dfrac{a^m}{a^n}=a^{m-n}[/tex] [tex]\implies \sf \dfrac{8^x}{2^y}=2^{(12+y-y)} \\\\ \implies \sf \dfrac{8^x}{2^y}=2^{12}[/tex] Therefore, the required value is 2¹² ________________________________ Laws of exponents : [tex]\dag \ \sf a^m \times a^n=a^{m+n} \\\\ \dag \ \sf \dfrac{a^m}{a^n}=a^{m-n} \\\\ \dag \ \sf (a^m)^n=a^{mn} \\\\ \dag \ \sf a^{-n}=\dfrac{1}{a^n} \\\\ \dag \ \sf \sqrt[n]{a} =a^{\dfrac{1}{n}}[/tex] Reply
Answer:
option (A) 2¹²
Step-by-step explanation:
Given :
3x – y = 12
To find :
the value of \sf \dfrac{8^x}{2^y}
2
y
8
x
Solution :
3x – y = 12
3x = 12 + y —[1]
Now, 8 can be written as (2 × 2 × 2)
i.e., 8 = 2³
\longrightarrow \sf 8^x = (2^3)^x = 2^{3x}⟶8
x
=(2
3
)
x
=2
3x
Put 3x = 12 + y [ ∵ eqn. 1 ]
\longrightarrow \sf 2^{3x}=2^{12+y}⟶2
3x
=2
12+y
So,
\begin{gathered}\implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{3x}}{2^y} \\\\ \implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{12+y}}{2^y}\end{gathered}
⟹
2
y
8
x
=
2
y
2
3x
⟹
2
y
8
x
=
2
y
2
12+y
We know, \rm \dfrac{a^m}{a^n}=a^{m-n}
a
n
a
m
=a
m−n
\begin{gathered}\implies \sf \dfrac{8^x}{2^y}=2^{(12+y-y)} \\\\ \implies \sf \dfrac{8^x}{2^y}=2^{12}\end{gathered}
⟹
2
y
8
x
=2
(12+y−y)
⟹
2
y
8
x
=2
12
Therefore, the required value is 2¹²
________________________________
Laws of exponents :
\begin{gathered}\dag \ \sf a^m \times a^n=a^{m+n} \\\\ \dag \ \sf \dfrac{a^m}{a^n}=a^{m-n} \\\\ \dag \ \sf (a^m)^n=a^{mn} \\\\ \dag \ \sf a^{-n}=\dfrac{1}{a^n} \\\\ \dag \ \sf \sqrt[n]{a} =a^{\dfrac{1}{n}\end{gathered}
Step-by-step explanation:
pls mark brainliest
Answer:
option (A) 2¹²
Step-by-step explanation:
Given :
3x – y = 12
To find :
the value of [tex]\sf \dfrac{8^x}{2^y}[/tex]
Solution :
3x – y = 12
3x = 12 + y —[1]
Now, 8 can be written as (2 × 2 × 2)
i.e., 8 = 2³
[tex]\longrightarrow \sf 8^x = (2^3)^x = 2^{3x}[/tex]
Put 3x = 12 + y [ ∵ eqn. 1 ]
[tex]\longrightarrow \sf 2^{3x}=2^{12+y}[/tex]
So,
[tex]\implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{3x}}{2^y} \\\\ \implies \sf \dfrac{8^x}{2^y}=\dfrac{2^{12+y}}{2^y}[/tex]
We know, [tex]\rm \dfrac{a^m}{a^n}=a^{m-n}[/tex]
[tex]\implies \sf \dfrac{8^x}{2^y}=2^{(12+y-y)} \\\\ \implies \sf \dfrac{8^x}{2^y}=2^{12}[/tex]
Therefore, the required value is 2¹²
________________________________
Laws of exponents :
[tex]\dag \ \sf a^m \times a^n=a^{m+n} \\\\ \dag \ \sf \dfrac{a^m}{a^n}=a^{m-n} \\\\ \dag \ \sf (a^m)^n=a^{mn} \\\\ \dag \ \sf a^{-n}=\dfrac{1}{a^n} \\\\ \dag \ \sf \sqrt[n]{a} =a^{\dfrac{1}{n}}[/tex]