Question Find the range of $p$ for which the simultaneous equations $p 2^{x} + \log_{2} y = 2$ and $2^{x} + \log_{2} y = 1$ have a real solution $(x, y)$ .

Studdy Solution

STEP 1

Assumptions1. The equations given are $\begin{array}{l} p^{x}+\log{} y= \\ ^{x}+\log{} y=1\end{array}$ . $is a real number3. We are looking for the range of$ for which these simultaneous equations have a real solution $(x, y)$

STEP 2

First, we can subtract the second equation from the first to eliminate $\log{2} y$ and simplify the equations.
$p2^{x}+\log{2} y - (2^{x}+\log{2} y) =2 -1$

STEP 3

This simplifies to $(p -1)2^{x} =1$

STEP 4

We can then solve for $2^{x}$ $2^{x} = \frac{1}{p -1}$

STEP 5

Since $2^{x}$ is always positive for real $x$ , $\frac{1}{p -1}$ must also be positive. This gives us the inequality $\frac{1}{p -1} >0$

STEP 6

We can solve this inequality by considering two cases $p -1 >0$ and $p -1 <0$ .

STEP 7

If $p -1 >0$ , then $p >1$ .

STEP 8

If $p -1 <0$ , then $p <1$ .

STEP 9

However, if $p <$ , then $\frac{}{p -}$ is negative, which contradicts the fact that $2^{x}$ is always positive.

STEP 10

Therefore, the only valid solution is $p >$ .
The complete range of $p$ for which these simultaneous equations have a real solution $(x, y)$ is $p >$ .

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Question Find the range of $p$ for which the simultaneous equations $p 2^{x} + \log_{2} y = 2$ and $2^{x} + \log_{2} y = 1$ have a real solution $(x, y)$ .

Studdy Solution

STEP 1

Assumptions1. The equations given are $\begin{array}{l} p^{x}+\log{} y= \\ ^{x}+\log{} y=1\end{array}$ . $is a real number3. We are looking for the range of$ for which these simultaneous equations have a real solution $(x, y)$

STEP 2

First, we can subtract the second equation from the first to eliminate $\log{2} y$ and simplify the equations.
$p2^{x}+\log{2} y - (2^{x}+\log{2} y) =2 -1$

STEP 3

This simplifies to $(p -1)2^{x} =1$

STEP 4

We can then solve for $2^{x}$ $2^{x} = \frac{1}{p -1}$

STEP 5

Since $2^{x}$ is always positive for real $x$ , $\frac{1}{p -1}$ must also be positive. This gives us the inequality $\frac{1}{p -1} >0$

STEP 6

We can solve this inequality by considering two cases $p -1 >0$ and $p -1 <0$ .

STEP 7

If $p -1 >0$ , then $p >1$ .

STEP 8

If $p -1 <0$ , then $p <1$ .

STEP 9

However, if $p <$ , then $\frac{}{p -}$ is negative, which contradicts the fact that $2^{x}$ is always positive.

STEP 10

Therefore, the only valid solution is $p >$ .
The complete range of $p$ for which these simultaneous equations have a real solution $(x, y)$ is $p >$ .

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Question Find the range of ppp for which the simultaneous equations p2x+log⁡2y=2p 2^{x} + \log_{2} y = 2p2x+log2​y=2 and 2x+log⁡2y=12^{x} + \log_{2} y = 12x+log2​y=1 have a real solution (x,y)(x, y)(x,y).

Studdy Solution

STEP 1

STEP 2

First, we can subtract the second equation from the first to eliminate log⁡2y\log{2} ylog2y and simplify the equations.p2x+log⁡2y−(2x+log⁡2y)=2−1p2^{x}+\log{2} y - (2^{x}+\log{2} y) =2 -1p2x+log2y−(2x+log2y)=2−1

STEP 3

This simplifies to(p−1)2x=1(p -1)2^{x} =1(p−1)2x=1

STEP 4

We can then solve for 2x2^{x}2x2x=1p−12^{x} = \frac{1}{p -1}2x=p−11​

STEP 5

Since 2x2^{x}2x is always positive for real xxx, 1p−1\frac{1}{p -1}p−11​ must also be positive. This gives us the inequality1p−1>0\frac{1}{p -1} >0p−11​>0

STEP 6

We can solve this inequality by considering two cases p−1>0p -1 >0p−1>0 and p−1<0p -1 <0p−1<0.

STEP 7

If p−1>0p -1 >0p−1>0, then p>1p >1p>1.

STEP 8

If p−1<0p -1 <0p−1<0, then p<1p <1p<1.

STEP 9

However, if p<p <p<, then p−\frac{}{p -}p−​ is negative, which contradicts the fact that 2x2^{x}2x is always positive.

STEP 10

Therefore, the only valid solution is p>p >p>.The complete range of ppp for which these simultaneous equations have a real solution (x,y)(x, y)(x,y) is p>p >p>.

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Question Find the range of $p$ for which the simultaneous equations $p 2^{x} + \log_{2} y = 2$ and $2^{x} + \log_{2} y = 1$ have a real solution $(x, y)$ .

First, we can subtract the second equation from the first to eliminate $\log{2} y$ and simplify the equations.
$p2^{x}+\log{2} y - (2^{x}+\log{2} y) =2 -1$

This simplifies to $(p -1)2^{x} =1$

We can then solve for $2^{x}$ $2^{x} = \frac{1}{p -1}$

Since $2^{x}$ is always positive for real $x$ , $\frac{1}{p -1}$ must also be positive. This gives us the inequality $\frac{1}{p -1} >0$

We can solve this inequality by considering two cases $p -1 >0$ and $p -1 <0$ .

If $p -1 >0$ , then $p >1$ .

If $p -1 <0$ , then $p <1$ .

However, if $p <$ , then $\frac{}{p -}$ is negative, which contradicts the fact that $2^{x}$ is always positive.

Therefore, the only valid solution is $p >$ .
The complete range of $p$ for which these simultaneous equations have a real solution $(x, y)$ is $p >$ .