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# Solve the following equation:${2^{2x + 1}} - 33 \cdot {2^{x - 1}} + 4 = 0$ Verified
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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.

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