Question

$\Large{\text{The equation: }} \displaystyle x^{3/4(log_2 \ x)^2 +(log_2 \ x) -5/4} = 2 \ has;$

a) at least one real solution
b) exactly one irrational solution
c) exactly three real solutions
d) all of the above

Answer 1

Answer:

$x^{3/4(log_2x)^2+log_2x-5/4}=\sqrt{2}\\\\\textbf{Taking log_2 both sides we get :}\\\\\implies log_2[x^{3/4(log_2x)^2+log_2x-5/4}]=log_2\sqrt{2}$

$[3/4(log_2x)^2+log_2x-5/4]log_2x=\dfrac{1}{2}log_22\\\\\implies [3/4(log_2x)^2+log_2x-5/4]log_2x=\dfrac{1}{2}\\\\\textbf{Take log_2x = k}\\\\\implies [3/4k^2+k-5/4]k=\dfrac{1}{2}\\\\\implies \dfrac{3}{4}k^3+k^2-5k=\dfrac{1}{2}\\\\\implies 3k^3+4k^2-5k-2=0\\\\\implies 3k^3+6k^2-2k^2-4k-k-2=0\\\\\implies 3k(k+2)-2k(k+2)-1(k+2)=0$

$\implies (3k^2-2k-1)(k+2)=0\\\\\implies (3k^2-3k+k-1)(k+2)=0\\\\\implies (3k+1)(k+2)(k-1)=0\\\\\implies k=-\dfrac{1}{3},1,-2$

$log_2x=1,-2,-1/3\\\\\implies x=2,\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{4}$

There are three real solutions .

Answer 2

Hope it helps ! mark as brainliest mistake in R.H.S ....√2