Question

(1 point) Solve for x (square root lnx)/(ln(square root x))Note, there are 2 possible solutions, A and B, where A < B.A= Is A a solution (yes or no)?B=Is B a solution (yes or no?

Answer 1

Problem Statement

The question asks us to solve the question given below:

$\frac{\sqrt[]{\ln x}}{\ln \sqrt[]{x}}=6$

Solution

$\begin{gathered} \frac{\sqrt[]{\ln x}}{\ln\sqrt[]{x}}=6 \\ \text{ Cross multiply} \\ \sqrt[]{\ln x}=6\times\ln \sqrt[]{x} \\ \text{Use the law of logarithm that says:} \\ a\ln b=\ln b^a,\text{ to the right-hand side} \\ \\ \sqrt[]{\ln x}=\ln (\sqrt[]{x})^6 \\ \sqrt[]{x}=x^{\frac{1}{2}} \\ \\ \sqrt[]{\ln x}=\ln x^{\frac{6}{2}}=\ln x^3 \\ \text{Applying the law stated above to the right-hand side again.} \\ \\ \sqrt[]{\ln x}=3\ln x \\ \text{square both sides} \\ (\sqrt[]{\ln x^{}})^2=(3\ln x)^2 \\ \\ \ln x=3^2(\ln x)^2=9(\ln x)^2 \\ \text{subtract }lnx\text{ from both sides} \\ 9(\ln x)^2-\ln x=0 \\ \text{Factorize} \\ \ln x(9\ln x-1)=0 \\ \therefore\ln x=0\text{ or} \\ 9\ln x-1=0 \\ \\ \therefore\ln x=0,x=e^0=1 \\ \\ 9\ln x=1 \\ \ln x=\frac{1}{9} \\ x=e^{\frac{1}{9}} \\ \\ \text{The values of x are:} \\ x\text{ = 1 or x}=e^{\frac{1}{9}} \end{gathered}$