Question

i need help for this question plzz​

Answer 1

Answer:

x=2

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Answer 2

Question:-

If $\sf {4} ^{x - 2} - {2}^{x + 1} = 0$ then the value of x is

Answer:-

Given:

$\sf \large{ {4}^{x - 2} - {2}^{x + 1} = 0}$

Transposing - (2) ^ (x + 1) to RHS we get,

$\sf \implies\large{ {4}^{x - 2} = {2}^{x + 1} }$

4 can be written as 2².

Hence,

$\sf \implies\large{ { ({2}^{2}) }^{x - 2} = {2}^{x + 1} }$

Using the formula [(a) ^ m]^n = a^(mn) in LHS we get,

$\sf \implies \large{ {2}^{2x - 4} = {2}^{x + 1} }$

Bases are equal hence the powers are also equal.

→ 2x - 4 = x + 1

→ 2x - x = 1 + 4

→ x = 5

Hence, the value of x is 5.

Some Important Formulae:

• $\sf a^m \times a^n = {a}^{m + n}$

• $\sf a^m \div a^n = {a}^{m - n}$

• $\sf \dfrac{1}{a^m} = {a}^{- m}$

• $\sf a^m \times b^m = {(ab)}^{m}$

• $\sf {\bigg(\dfrac{a}{b}\bigg)}^{ - n} = {\bigg(\dfrac{b}{a}\bigg)}^{ n}$

• a⁰ = 1

• a¹ = a