# Solving an equation with logarithms: $x^{\log_2(\sqrt{x})-1} = \sqrt{8}$

The equation I'm given is $$\large x^{\log_2(\sqrt{x})-1} = \sqrt{8}$$

I've tried on solving it and my best try is on the photo. Got stuck there and not sure how to proceed any further.

• Looking over your work, for your last equation, substitute $u = \log_{2}(x)$ so you have a quadratic. – MathematicsStudent1122 Jun 11 '16 at 17:11
• what about solving the second degree equation in $\log_2 x$ ? if you like it change the variable as $y=\log_2 x$ and solve $y^2-2y-3=0$ you'll get two solutions say $y_1$ and $y_2$ and then solve $\log_2 x=y_{1;2}$ – Renato Faraone Jun 11 '16 at 17:11
• Yes, actually Overkillus has already done the harder part of the work! He only need to solve tha standard quadratic equation – guestDiego Jun 11 '16 at 17:17

$${ x }^{ \log _{ 2 }{ \left( \sqrt { x } \right) -1 } }=\sqrt { 8 } \\ \log _{ 2 }{ \left( { x }^{ \log _{ 2 }{ \left( \sqrt { x } \right) -1 } } \right) =\log _{ 2 }{ { 2 }^{ \frac { 3 }{ 2 } } } } \\ \left( \log _{ 2 }{ \left( \sqrt { x } \right) -1 } \right) \log _{ 2 }{ x } =\frac { 3 }{ 2 } \\ \left( \frac { 1 }{ 2 } \log _{ 2 }{ x } -1 \right) \log _{ 2 }{ x } =\frac { 3 }{ 2 } \\ \text{Let } \log _{ 2 }{ x } =t \text{ and substitute.}\\ { t }^{ 2 }-2t-3=0\\ \left( t+1 \right) \left( t-3 \right) =0\\ \log _{ 2 }{ x } =-1\Rightarrow x=0.5\\ \log _{ 2 }{ x } =3\Rightarrow x=8$$

• domain shoud be positive since $\sqrt { x } >0\Rightarrow x>0$ so both are solution of equation – haqnatural Jun 11 '16 at 18:17

$$x^{\log_2(\sqrt{x})-1} = \sqrt{8}$$ $$\log_2(x^{\log_2(\sqrt{x})-1})=\log_2(\sqrt{8})$$ $$(\log_2(\sqrt{x})-1)\log_2(x)=\frac{1}{2}\log_2(8)$$ $$\frac{1}{2}\log_2(8)\Rightarrow\frac{3}{2}$$ $$\left(\frac{1}{2}\log_2(x)-1\right)\log_2(x)=\frac{3}{2}$$ $$\log_2(x)=u$$ $$\left(\frac{1}{2}u-1\right)u*2=\frac{3}{2}*2$$ $$u^2-2u-3=0$$ $$u=\frac{2+\sqrt{(-2)^2-4*1*(-3)}}{2}=3,u=\frac{2+\sqrt{(-2)^2-4*1*(-3)}}{2}=-1$$ $$\log_2(x)=3\Rightarrow x=8,\log_2(x)=-1\Rightarrow x=\frac{1}{2}$$ $$8^{\log_2(\sqrt{8})-1} = \sqrt{8},\frac{1}{2}^{\log_2(\sqrt{\frac{1}{2}})-1} = \sqrt{8}$$

Fill in details and be sure you understand what properties are being applied:

$$x^{\log_2\sqrt x-1}=\sqrt8\implies\left(\frac12\log_2x-1\right)\log_2x=\frac32\implies$$

$$\left(\log_2x\right)^2-2\log_2x-3=0\implies\left(\log_2x-3\right)\left(\log_2x+1\right)=0$$

Finish now the exercise...and observe the only really new thing from what you did is the very last step

$$\large\log_2\sqrt{x}=t\implies x=2^{2t}$$ $$\large{2^{2t(t-1)}}=2^{\frac{3}{2}} \implies 4t^2-4t-3=0\implies t_1=-\frac{1}{2}\,,\, t_2=\frac{3}{2}$$ then $$\large x_1=\frac{1}{2}\,\,,\,\, x_2=8$$