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Question: The equation \({x^{\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}} =...

The equation x34(log⁡2x)2+log⁡2x−54=2{x^{\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}} = \sqrt 2 has
A. at least one real solution
B. exactly three real solutions
C. exactly one irrational solution
D. complex roots

Explanation

Solution

Simplify the given expression by taking log both sides. Then, substitute log⁡2x=t{\log _2}x = t and find the factor of the equation by hit and trial method. Then, write all the factors of the equation. Next, equate each of the factors to 0 to find the value of tt and hence the value of xx.

Complete step-by-step answer:
We will solve the equation x34(log⁡2x)2+log⁡2x−54=2{x^{\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}} = \sqrt 2
Let us simplify the equation by taking log with base 2 on both sides.
log⁡2(x34(log⁡x)2+log⁡2x−54)=log⁡22{\log _2}\left( {{x^{\dfrac{3}{4}{{\left( {\log x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}}} \right) = {\log _2}\sqrt 2
But, we know that log⁡abn=nlog⁡ab{\log _a}{b^n} = n{\log _a}b
(34(log⁡2x)2+log⁡2x−(54))log⁡2x=log⁡22\left( {\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \left( {\dfrac{5}{4}} \right)} \right){\log _2}x = {\log _2}\sqrt 2
Let log⁡2x=t{\log _2}x = t
(34t2+t−54)t=12\left( {\dfrac{3}{4}{t^2} + t - \dfrac{5}{4}} \right)t = \dfrac{1}{2}
On cross-multiplying and simplifying the equation , we will get,
34t3+t2−54t=12 ⇒32t3+2t2−52t=1 ⇒3t3+4t2−5t−2=0  \dfrac{3}{4}{t^3} + {t^2} - \dfrac{5}{4}t = \dfrac{1}{2} \\\ \Rightarrow \dfrac{3}{2}{t^3} + 2{t^2} - \dfrac{5}{2}t = 1 \\\ \Rightarrow 3{t^3} + 4{t^2} - 5t - 2 = 0 \\\
We will use hit and trial to find the root of the above equation,
Put t=1t = 1 in the above equation,
3(1)3+4(1)2−5(1)−2 =3+4−5−2 =7−7 =0  3{\left( 1 \right)^3} + 4{\left( 1 \right)^2} - 5\left( 1 \right) - 2 \\\ = 3 + 4 - 5 - 2 \\\ = 7 - 7 \\\ = 0 \\\
Hence, t−1t - 1 is the factor of the equation, 3t3+4t2−5t−2=03{t^3} + 4{t^2} - 5t - 2 = 0
Therefore, we can divide the equation by t−1t - 1 and write the equation as
(t−1)(3t2+7t+2)=0\left( {t - 1} \right)\left( {3{t^2} + 7t + 2} \right) = 0
Factorise the above expression,
(t−1)(3t2+6t+t+2)=0 ⇒(t−1)(3t(t+2)+1(t+2))=0 ⇒(t−1)(3t+1)(t+2)=0  \left( {t - 1} \right)\left( {3{t^2} + 6t + t + 2} \right) = 0 \\\ \Rightarrow \left( {t - 1} \right)\left( {3t\left( {t + 2} \right) + 1\left( {t + 2} \right)} \right) = 0 \\\ \Rightarrow \left( {t - 1} \right)\left( {3t + 1} \right)\left( {t + 2} \right) = 0 \\\
Equate each factor to 0 and the value of tt
t−1=0 ⇒t=1  t - 1 = 0 \\\ \Rightarrow t = 1 \\\
3t+1=0 ⇒t=−13  3t + 1 = 0 \\\ \Rightarrow t = - \dfrac{1}{3} \\\
And
t+2=0 ⇒t=−2  t + 2 = 0 \\\ \Rightarrow t = - 2 \\\
Substitute back t=log⁡2xt = {\log _2}x and hence find the value of xx
t=1 ⇒log⁡2x=1 ⇒x=21 ⇒x=2  t = 1 \\\ \Rightarrow {\log _2}x = 1 \\\ \Rightarrow x = {2^1} \\\ \Rightarrow x = 2 \\\
Similarly,
t=−2 ⇒log⁡2x=−2 ⇒x=2−2 ⇒x=14  t = - 2 \\\ \Rightarrow {\log _2}x = - 2 \\\ \Rightarrow x = {2^{ - 2}} \\\ \Rightarrow x = \dfrac{1}{4} \\\
And
t=−13 ⇒log⁡2x=−13 ⇒x=2−13  t = - \dfrac{1}{3} \\\ \Rightarrow {\log _2}x = - \dfrac{1}{3} \\\ \Rightarrow x = {2^{ - \dfrac{1}{3}}} \\\
Since, xx has three real values.
Therefore, there are exactly three real solutions for the given equation.
Hence, option C is correct.

Note: If the number is written of the form log⁡ab=x{\log _a}b = x, then aa is the base of the logarithmic function and b=axb = {a^x}. Also, students must know the properties of logarithmic function, like log⁡abn=nlog⁡ab{\log _a}{b^n} = n{\log _a}b and log⁡aan=n{\log _a}{a^n} = n