Question

can any one solve this question ??​

Answer 1

Question :

If $\sf 4^x - 4^{x-1}=24$ , then find the value of x .

Solution :

$\sf 4^x - 4^{x-1}=24$

$\bullet\ \; \sf \pink{a^{m+n}=a^m . a^n}$

$\bullet\ \; \sf \pink{a^{m-n}=\dfrac{a^m}{a^n}}$

$\bullet\ \; \sf \pink{a^{-1}=\dfrac{1}{a}}$

$:\implies \sf 4^x - 4^x .4^{-1} =24$

$:\implies \sf 4^x - \dfrac{4^x }{4} =24$

$:\implies \sf 4^x \left( 1- \dfrac{1}{4}\right) =24$

$:\implies \sf 4^x \left( \dfrac{3}{4}\right) =24$

$:\implies \sf 4^x=32$

$:\implies \sf \left(2^2\right)^x = (2)^5$

$:\implies \sf (2)^{2x}=(2)^5$

• Bases are equal , so exponents must be equal .

$:\implies \sf 2x=5$

$:\implies \green{ \textsf{\textbf{x = }} \dfrac{\textsf{\textbf{5}}}{\textsf{\textbf{2}}}}\ \; \; \bigstar$

Answer 2

Answer:

Given :-

• $\sf {4}^{x} -\: {4}^{x - 1} =\: 24$

To Find :-

• What is the value of x.

Formula Used :-

$\leadsto$ $\sf {a}^{m} \times {a}^{n} =\: {a}^{m + n}$

$\leadsto$ $\sf \dfrac{{a}^{m}}{{a}^{n}} =\: {a}^{m - n}$

$\leadsto$ $\sf {a}^{- m} =\: \dfrac{1}{{a}^{m}}$

Solution :-

$\longmapsto$ $\sf {4}^{x} -\: {4}^{x - 1} =\: 24$

$\implies$ $\sf {4}^{x} -\: {4}^{x} \times {4}^{- 1} =\: 24$

By taking 4x common we get,

$\implies$ $\sf {4}^{x}\bigg(1 - {4}^{- 1}\bigg) =\: 24$

$\implies$ $\sf {4}^{x}\bigg(1 - \dfrac{1}{4}\bigg) =\: 24$

$\implies$ $\sf {4}^{x}\bigg(\dfrac{4 - 1}{4}\bigg) =\: 24$

$\implies$ $\sf {4}^{x}\bigg(\dfrac{3}{4}\bigg) =\: 24$

$\implies$ $\sf {4}^{x} =\: 24 \times \dfrac{4}{3}$

$\implies$ $\sf {4}^{x} =\: \dfrac{\cancel{96}}{\cancel{3}}$

$\implies$ $\sf {4}^{x} =\: 32$

$\implies$ $\sf {\cancel{(2)}}^{2x} =\: {\cancel{(2)}}^{5}$

$\implies$ $\sf 2x =\: 5$

$\implies$ $\sf\bold{\purple{x =\: \dfrac{5}{2}}}$

${\underline{\boxed{\small{\bf{\therefore The\: value\: of\: x\: is\: \dfrac{5}{2}\: .}}}}}$

SOME IMPORTANT FORMULA :-

↦ $\sf ({a}^{m})^{n} =\: {a}^{m \times n}$

↦ $\sf \bigg({\dfrac{a}{b}}\bigg)^{m} =\: \dfrac{a^m}{b^m}$

↦ $\sf {a}^{1} =\: a$

↦ $\sf {a}^{0} =\: 1$

↦ $\sf {a}^{\dfrac{x}{y}} =\: \sqrt[y]{{a}^{x}}$