Question

Find the value which satisfies the expression 4^x - 3.2^(x+3)+128 = 0

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given an equation
• $\sf{4^x - 3.2^{x+3}+128} = 0$

To Find:

• We have to find the values of x which satisfies the given equation

Solution:

We have been given an equation

$\hookrightarrow \sf{4^x - 3(2^{x+3}) +128 = 0}$

Using law of exponents

$\boxed{\sf{\orange{a^{m+n} = a^m \times a^n}}}$

$\hookrightarrow \sf{4^x - 3(2^x)(2^3) + 128 = 0}$

$\hookrightarrow \sf{{(2^x)}^{2}-24(2^x)+ 128 = 0}$

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$\bigstar \: \: \underline{\LARGE\mathfrak\pink{Let \: us \: assume:}}$

$\boxed{\sf{\red{2^x = y }}}$ --------------( 1 )

Hence the equation becomes

$\hookrightarrow \sf{ y^2 - 24y + 128 = 0}$

Solving the equation using middle term splitting

$\hookrightarrow \sf{y^2 - 16y - 8y + 128 = 0}$

$\hookrightarrow \sf{y( y - 16 ) - 8 ( y - 16 ) = 0}$

$\hookrightarrow \sf{(y - 16 ) ( y - 8 ) = 0}$

Hence Either

$\implies \sf{ y - 16 = 0 }$

$\implies \sf{y = 16}$

Or

$\implies \sf{ y - 8 = 0 }$

$\implies \sf{ y = 8 }$

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Putting values of y in ( 1 )

1 ) When y = 16

$\longrightarrow \sf{2^x = 16}$

$\longrightarrow \sf{2^x = 2^4}$

Comparing the bases on both sides of equation

$\longrightarrow \boxed{\sf{x = 4}}$

2 ) When y = 8

$\longrightarrow \sf{2^x = 8}$

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

Comparing the bases on both sides of equation

$\longrightarrow \boxed{\sf{ x = 3}}$

$\boxed{\sf{\red{Value \: of \: x = 3 \: or \: 4}}}$

Hence value of x = 3 or x = 4 satisfies the given equation

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$\huge\mathfrak\green{Verification:}$

1 ) When x = 3

$\implies \sf{4^3 - 3.2^{3+3}+128}$

$\implies \sf{64 - 3.2^6 + 128}$

$\implies \sf{192 - 3 \times 64}$

$\implies \sf{192 - 192 = 0 \: ( Answer ) }$

2 ) When x = 4

$\implies \sf{4^4 - 3.2^{4+3} + 128 }$

$\implies \sf{256 - 3.2^7 + 128}$

$\implies \sf{384 - 3 \times 128}$

$\implies \sf{384 - 384 = 0 \: (Answer ) }$

Hence Verified !!

Answer 2

Step-by-step explanation:

$\huge\mathfrak\red{Answer}$

Given Equation;

$4x - 3( {2}^{x + 3} + (128) = 0$

Using law of exponents:

${a}^{m + n} = {a}^{m} \times {a}^{n}$

$= > ( {2}^{2} ) ^{x} - 3( {2}^{x} . {2}^{3} + 128 = 0$

$= > ( {2}^{x} ) ^{2} - 3( {2}^{x}. 8) + 128 = 0$

let us assume that:

${2}^{x} = y$

The Equation become

${y}^{2} - 24y + 128 = 0$

Splitting middle term

$= > {y}^{2} - 16y - 8y + 128 = 0$

$= > y(y - 16) - 8(y - 16) = 0$

$(y - 16)(y - 8) = 0$

So, y=8 and you=16

$when \: y = 8 = > {2}^{x} = {2}^{3} = > x = 3$

$when \: y = 16 = > {2}^{x} = {2}^{4} = > x = 4$

Hence roots are 3 and 4 .

So, the sum of the roots =3+4=7