Question

Exponent
Can you please solve question c. Thank you

Answer 1

Solution

$\tt \to \: 8 \times {2}^{2x} + 4 \times {2}^{x + 1} = 1 + {2}^{x}$

Now we can write as

$\tt \to \: 8 \times {2}^{2x} + 4 \times {2}^{x} \times 2 = 1 + {2}^{x}$

$\tt \to \: 8 \times {2}^{2x} + 8 \times {2}^{x} = 1 + {2}^{x}$

Now let

$\to \tt \: {2}^{x} = y$

We get

$\tt \to \: 8 {y}^{2} + 8y = 1 + y$

$\tt \to \: 8 {y}^{2} + 7y - 1 = 0$

$\tt \to \: 8 {y}^{2} + 8y - y - 1 = 0$

$\tt \to \: 8y(y + 1) - 1(y + 1) = 0$

$\tt \to \: (8y - 1)(y + 1) = 0$

So we get

$\tt \to \: y = \dfrac{1}{8} \: and \: - 1$

so -1 is not possible because 2ˣ is always positive so we take 1/8

Now put the value y = 1/8

$\to \tt \: {2}^{x} = y$

$\to \tt \: {2}^{x} = \dfrac{1}{8}$

$\tt \to {2}^{x} = \dfrac{1}{ {2}^{3} }$

$\tt \to \: {2}^{x} = {2}^{ - 3}$

$\tt \to \: x = - 3$

Answer is - 3

Answer 2

Answer:

$\huge\bf\blue{\underline{\underline{AnsWeR}}}$

Solution

\tt \to \: 8 \times {2}^{2x} + 4 \times {2}^{x + 1} = 1 + {2}^{x}→8×2

2x

+4×2

x+1

=1+2

x

Now we can write as

\tt \to \: 8 \times {2}^{2x} + 4 \times {2}^{x} \times 2 = 1 + {2}^{x}→8×2

2x

+4×2

x

×2=1+2

x

\tt \to \: 8 \times {2}^{2x} + 8 \times {2}^{x} = 1 + {2}^{x}→8×2

2x

+8×2

x

=1+2

x

Now let

\to \tt \: {2}^{x} = y→2

x

=y

We get

\tt \to \: 8 {y}^{2} + 8y = 1 + y→8y

2

+8y=1+y

\tt \to \: 8 {y}^{2} + 7y - 1 = 0→8y

2

+7y−1=0

\tt \to \: 8 {y}^{2} + 8y - y - 1 = 0→8y

2

+8y−y−1=0

\tt \to \: 8y(y + 1) - 1(y + 1) = 0→8y(y+1)−1(y+1)=0

\tt \to \: (8y - 1)(y + 1) = 0→(8y−1)(y+1)=0

So we get

\tt \to \: y = \dfrac{1}{8} \: and \: - 1→y=

8

1

and−1

so -1 is not possible because 2ˣ is always positive so we take 1/8

Now put the value y = 1/8

\to \tt \: {2}^{x} = y→2

x

=y

\to \tt \: {2}^{x} = \dfrac{1}{8}→2

x

=

8

1

\tt \to {2}^{x} = \dfrac{1}{ {2}^{3} }→2

x

=

2

3

1

\tt \to \: {2}^{x} = {2}^{ - 3}→2

x

=2

−3

\tt \to \: x = - 3→x=−3

Answer is - 3