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Solve and find the domain of the equation:

 

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

 Feb 12, 2018

Best Answer 

 #2
avatar+22172 
+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

 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)

 Feb 13, 2018
 #2
avatar+22172 
+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

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