Question

solve for x ! indices chapter class 9 icse ​

Answer 1

Answer:

-1/2

Step-by-step explanation:

2x+(x+1) = x

3x+1 = x

1 = x-3x

1 = -2x

x = -1/2

Hopefully it'll help......

Answer 2

✪ $\purple{\Huge{\underline{\underline{ \rm{Solution }}}}}$✪

$\sf8 \times {2}^{2x} + 4 \times {2}^{x + 1} = 1 + {2}^{x}$

$\sf{We \: know, \: {(a \times b)}^{2} } = {a}^{m} \times {b}^{m}$

$\sf{ \therefore{ {2}^{x + 1} = {2}^{x} \times {2}^{1} }}$

$\sf8 \times {2}^{2x} + 4 \times {2}^{x } \times {2}^{1} = 1 + {2}^{x}$

$\sf8 \times { ({2}^{x}) }^{2} + 4 \times {2}^{x } \times {2}^{1} - 1 + {2}^{x} = 0$

Taking $\sf {2}^{x}$ as common, we have:

$8 \times { ({2}^{x} )}^{2} + {2}^{x} (4 \times 1 \times 2 - 1) - 1 = 0$

$\sf8 \times { ({2}^{x}) }^{2} + {2}^{x} \times (8 - 1) - 1 = 0$

$\sf8 \times { {(2}^{x}) }^{2} + 7 ({2}^{x}) - 1 = 0$

Let $\sf y = {2}^{x}$

$\sf {8y}^{2} + 7y - 1 = 0$

$\sf{ {8y}^{2} + 8y - y - 1 = 0}$

$\sf{8y(y + 1) - 1(y + 1) = 0}$

$\sf{(8y - 1) \: (y + 1) = 0}$

Now,

$\sf{8y - 1 = 0} \: or \: y + 1 = 0$

$\sf8y = 1 \: or \: y = - 1$

$\sf y = \dfrac{1}{8} \: or \: y = - 1$

Putting value of $\sf {2}^{x}$ , we have:

$\sf{2}^{x} = \dfrac{1}{8} \: or \: {2}^{x} = - 1$

Here we will not consider $\sf {2}^{x} = - 1$ because it is not possible.

Therefore, here we will take:

$\sf {2}^{x} = \dfrac{1}{8}$

Now,

$\sf {2}^{x} = \dfrac{1}{ {2}^{3} }$

$\sf {2}^{x} = {2}^{ - 3}$

When the bases are same powers are equal:

$\sf{ \green{ \underline{ \boxed{ \boxed{ \sf{x = - 3}}}}}}$

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