# Solve the following equation:

${2^{2x + 1}} - 33 \cdot {2^{x - 1}} + 4 = 0$

Answer

Verified

381k+ views

Hint: Here we go through by breaking the exponents into two terms i.e. by writing ${2^{2x + 1}} = 2 \cdot {2^{2x}}$ and ${2^{x - 1}} = \dfrac{{{2^x}}}{2}$then let the term ${2^x}$ as t and form the quadratic equation. By solving the quadratic equation we find the value of t after that equate with ${2^x}$to get the answer.

Complete step-by-step answer:

${2^{2x + 1}} - 33 \cdot {2^{x - 1}} + 4 = 0$

Now break the terms ${2^{2x + 1}} = 2 \cdot {2^{2x}}$and ${2^{x - 1}} = \dfrac{{{2^x}}}{2}$ as we know the rule of exponents i.e.${a^m} \cdot {a^n} = {a^{m + n}}$ and \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\].

Now we write it as,

now let the term ${2^x}$ as t.

$2{t^2} - \dfrac{{33}}{2}t + 4 = 0$ here we see that it is forming a quadratic equation.

$4{t^2} - 33t + 8 = 0$

Now we have to make it in factor form.

$4{t^2} - 32t - t + 8 = 0$

$4t\left( {t - 8} \right) - 1\left( {t - 8} \right) = 0$

$

\left( {t - 8} \right)\left( {4t - 1} \right) = 0 \\

\therefore t = 8,t = \dfrac{1}{4} \\

$

Now as we have assumed ${2^x} = t$

So, ${2^x} = 8 \Rightarrow \therefore x = 3$ or ${2^x} = \dfrac{1}{4} \Rightarrow \therefore x = - 2$

Hence x=3 and x= -2 are the required answer.

Note: For solving this type of question you have to use the concept of exponent and to solve it further consider an exponent as a variable then it may become a quadratic equation as in this question and then you can solve it easily.

Complete step-by-step answer:

${2^{2x + 1}} - 33 \cdot {2^{x - 1}} + 4 = 0$

Now break the terms ${2^{2x + 1}} = 2 \cdot {2^{2x}}$and ${2^{x - 1}} = \dfrac{{{2^x}}}{2}$ as we know the rule of exponents i.e.${a^m} \cdot {a^n} = {a^{m + n}}$ and \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\].

Now we write it as,

now let the term ${2^x}$ as t.

$2{t^2} - \dfrac{{33}}{2}t + 4 = 0$ here we see that it is forming a quadratic equation.

$4{t^2} - 33t + 8 = 0$

Now we have to make it in factor form.

$4{t^2} - 32t - t + 8 = 0$

$4t\left( {t - 8} \right) - 1\left( {t - 8} \right) = 0$

$

\left( {t - 8} \right)\left( {4t - 1} \right) = 0 \\

\therefore t = 8,t = \dfrac{1}{4} \\

$

Now as we have assumed ${2^x} = t$

So, ${2^x} = 8 \Rightarrow \therefore x = 3$ or ${2^x} = \dfrac{1}{4} \Rightarrow \therefore x = - 2$

Hence x=3 and x= -2 are the required answer.

Note: For solving this type of question you have to use the concept of exponent and to solve it further consider an exponent as a variable then it may become a quadratic equation as in this question and then you can solve it easily.

Recently Updated Pages

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How many meters are there in a kilometer And how many class 8 maths CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a labelled sketch of the human eye class 12 physics CBSE