Question

Solve for x: (1/2)lnx-ln(3x^2 +2) = ln (1/5)

Answer 1

\(\normalsize\color{blue}{ (1/2)\ln x-\ln(3x^2 +2) = \ln (1/5)\LARGE\color{white}{ \rm │ }}\) like this ?

Answer 2

\(\normalsize\color{blue}{ (1/2)\ln x-\ln(3x^2 +2) = \ln (1/5)\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ \ln \sqrt{~x}~ -\ln(3x^2 +2) = \ln (1/5)\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ \ln[~~\sqrt{~x}/(3x^2 +2)~~] = \ln (1/5)\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ \sqrt{~x}/(3x^2 +2)~~] = 1/5\LARGE\color{white}{ \rm │ }}\)

Answer 3

now, cross multiply, \(\normalsize\color{blue}{ \sqrt{~x}/(3x^2 +2)= 1/5\LARGE\color{white}{ \rm │ }}\)

\(\normalsize\color{blue}{ 5\sqrt{~x}=3x^2 +2\LARGE\color{white}{ \rm │ }}\)

Answer 4

Then raise both sides to the second power,

\(\normalsize\color{blue}{ 25x=9x^4 +12x^2+4\LARGE\color{white}{ \rm │ }}\)

Answer 5

\(\normalsize\color{blue}{9x^4 +12x^2-25x+4)=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ (x-1)(9x^3 +9x^2+21x-4)=0\LARGE\color{white}{ \rm │ }}\)

Answer 6

x-1=0, x=1

Answer 7

Solve for x from \(\normalsize\color{blue}{9x^3 +9x^2+21x-4=0\LARGE\color{white}{ \rm │ }}\)