+0  
 
+3
96
2
avatar+598 

Solve and find the domain of the equation:

 

\(x^{\log _\sqrt{x} (x-2)}=9\)

michaelcai  Feb 12, 2018

Best Answer 

 #2
avatar+19508 
+1

Solve and find the domain of the equation:

\(\huge{ x^{\log_\sqrt{x} (x-2)}=9 }\)

 

1. Domain:

\(x-2 > 0 \\ x > 2 \)

\(\text{ $\{x \in R : x>2\}$ (assuming a function from reals to reals) } \)

 

2. Solve:

\(\begin{array}{|rcll|} \hline \large{ x^{\log_\sqrt{x} (x-2)} }& \large{=}& \large{9} \\ && \boxed{ \log_\sqrt{x} (x-2) = \dfrac{\ln(x-2)}{\ln(\sqrt{x})} \\ = \dfrac{\ln(x-2)}{\ln(x^{\frac12})}\\ = \dfrac{\ln(x-2)}{\frac12\ln(x)}\\ = \dfrac{2\ln(x-2)}{\ln(x)}} \\ \large{ x^{\dfrac{2\ln(x-2)}{\ln(x)}}}& \large{=}& \large{9} \\ && \boxed{\text{Formula: } a^b = e^{\ln(a^b)} = e^{b\ln(a)} } \\ && \boxed{ x^{\dfrac{2\ln(x-2)}{\ln(x)}} = e^{ \dfrac{2\ln(x-2)}{\ln(x)} \cdot \ln(x) } =e^{2\ln(x-2)} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \large{ e^{2\ln(x-2)} } & \large{=} & \large{9} \\ \large{ e^{\ln\left((x-2)^2\right)} } & \large{=} & \large{9} \quad & | \quad e^{\ln\left((x-2)^2\right)} = (x-2)^2 \\ \large{ (x-2)^2 } & \large{=} & \large{9} \quad & | \quad \sqrt{} \text{ both sides} \\ \large{ x-2 } & \large{=} & \large{\pm 3} \\\\ x_1 -2 &=& 3 \\ x_1 &=& 3+2 \\ \mathbf{x_1} & \mathbf{=} & \mathbf{5} \\\\ x_2 -2 &=& -3 \\ x_2 &=& -3+2 \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-1} \quad & | \quad \text{no solution, see domain } x> 2 \\ \hline \end{array}\)

 

laugh

heureka  Feb 13, 2018
 #1
avatar
+1

Solve for x:
x^(log(x)/(log(sqrt(x)))) = 9

Simplify and substitute y = sqrt(x).
x^(log(x)/(log(sqrt(x)))) = sqrt(x)^4
 = y^4:
y^4 = 9

Taking 4^th roots gives sqrt(3) times the 4^th roots of unity:
y = -sqrt(3) or y = -i sqrt(3) or y = i sqrt(3) or y = sqrt(3)

Substitute back for y = sqrt(x):
sqrt(x) = -sqrt(3) or sqrt(x) = -i sqrt(3) or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or sqrt(x) = -i sqrt(3) or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or x = -3 or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or x = -3 or x = -3 or sqrt(x) = sqrt(3)
Raise both sides to the power of two:
x = 3    or x = -3     or x = -3      or x = 3

 

Domain: {x element R : 2 3 + sqrt(5)} (assuming a function from reals to reals)

Guest Feb 13, 2018
 #2
avatar+19508 
+1
Best Answer

Solve and find the domain of the equation:

\(\huge{ x^{\log_\sqrt{x} (x-2)}=9 }\)

 

1. Domain:

\(x-2 > 0 \\ x > 2 \)

\(\text{ $\{x \in R : x>2\}$ (assuming a function from reals to reals) } \)

 

2. Solve:

\(\begin{array}{|rcll|} \hline \large{ x^{\log_\sqrt{x} (x-2)} }& \large{=}& \large{9} \\ && \boxed{ \log_\sqrt{x} (x-2) = \dfrac{\ln(x-2)}{\ln(\sqrt{x})} \\ = \dfrac{\ln(x-2)}{\ln(x^{\frac12})}\\ = \dfrac{\ln(x-2)}{\frac12\ln(x)}\\ = \dfrac{2\ln(x-2)}{\ln(x)}} \\ \large{ x^{\dfrac{2\ln(x-2)}{\ln(x)}}}& \large{=}& \large{9} \\ && \boxed{\text{Formula: } a^b = e^{\ln(a^b)} = e^{b\ln(a)} } \\ && \boxed{ x^{\dfrac{2\ln(x-2)}{\ln(x)}} = e^{ \dfrac{2\ln(x-2)}{\ln(x)} \cdot \ln(x) } =e^{2\ln(x-2)} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \large{ e^{2\ln(x-2)} } & \large{=} & \large{9} \\ \large{ e^{\ln\left((x-2)^2\right)} } & \large{=} & \large{9} \quad & | \quad e^{\ln\left((x-2)^2\right)} = (x-2)^2 \\ \large{ (x-2)^2 } & \large{=} & \large{9} \quad & | \quad \sqrt{} \text{ both sides} \\ \large{ x-2 } & \large{=} & \large{\pm 3} \\\\ x_1 -2 &=& 3 \\ x_1 &=& 3+2 \\ \mathbf{x_1} & \mathbf{=} & \mathbf{5} \\\\ x_2 -2 &=& -3 \\ x_2 &=& -3+2 \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-1} \quad & | \quad \text{no solution, see domain } x> 2 \\ \hline \end{array}\)

 

laugh

heureka  Feb 13, 2018

15 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.