Question

$\sf {a^{\frac{3}{2} [log_{2}(a) - 3]} = \frac{1}{8}} \\\\ \textsf{then the value of a is}$



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Answer 1

$\begin{aligned}&\mathsf{a^{\frac{3}{2}[\log_2(a)-3]}=\dfrac{1}{8}}\\\\\implies\ \ &\mathsf{a^{\frac{3}{2}[\log_2(a)-3]}=8^{-1}}\\\\\implies\ \ &\mathsf{a^{\frac{3}{2}[\log_2(a)-3]}=(2^3)^{-1}}\\\\\implies\ \ &\mathsf{2^{\frac{3}{2}\log_2(a)[\log_2(a)-3]}=2^{-3}\quad\quad\left[\because\ a^b=m^{b\log_m(a)}\right]}\end{aligned}$

Since the base is 2, we can equate the powers.

$\begin{aligned}&\mathsf{\dfrac{3}{2}\log_2(a)[\log_2(a)-3]=-3}\\\\\implies\ \ &\mathsf{\log_2(a)[\log_2(a)-3]=-2}\\\\\implies\ \ &\mathsf{(\log_2(a))^2-3\log_2(a)+2=0}\\\\\implies\ \ &\mathsf{(\log_2(a))^2-\log_2(a)-2\log_2(a)+2=0}\\\\\implies\ \ &\mathsf{\log_2(a)(\log_2(a)-1)-2(\log_2(a)-1)=0}\\\\\implies\ \ &\mathsf{(\log_2(a)-1)(\log_2(a)-2)=0}\\\\\implies\ \ &\mathsf{\log_2(a)=1\quad;\quad\log_2(a)=2}\\\\\implies\ \ &\mathbf{a=2\quad\quad\ \ \ \ ;\ \ \ \ \quad\quad a=4}\end{aligned}$