Any idea how to solve this equation? \(\displaystyle{x}^{{2}}{{\log}_{{{3}}}{x}^{{2}}}-{\left({2}{x}^{{2}}+{3}\right)}{{\log}_{{{9}}}{\left({2}{x}+{3}\right)}}={3}{{\log}_{{{3}}}{\frac{{{x}}}{{{2}{x}+{3}}}}}\)

Carlo Sawyer

Carlo Sawyer

Answered question

2022-03-18

Any idea how to solve this equation?
x2log3x2(2x2+3)log9(2x+3)=3log3x2x+3

Answer & Explanation

zonadeenfoqueun6

zonadeenfoqueun6

Beginner2022-03-19Added 3 answers

Note that log9(x)=log3(x)log3(9)=12log3(x)
We need 2x+3>0, and hence x>0, so that all logarithms can be evaluated.
So we can simplify the equation using the basic properties of the logarithms
x2log3x2(2x2+3)log9(2x+3)=3log3(x2x+3)
2x2log3(x)2x2+32log3(2x+3)=3log3(x)3log3(2x+3)
(2x23)log3(x)=(2x2+323)log3(2x+3)
(2x23)log3(x)=2x232log3(2x+3)
From this, it is clear that any solutions to 2x23=0 will satisfy the equation. The solutions are x=32 and x=32 but the latter cannot be used in the original equation. So one possibility is
x=32.
If 2x230 , then cancelling we get:
(2x23)log3(x)=2x232log3(2x+3)
log3(x)=12log3(2x+3)
2log3(x)log3(2x+3)=0
log3(x22x+3)=0
x22x+3=1
x22x3=0
(x3)(x+1)=0
Again, x=1 cannot be used. So the only other solution is x=3
So the solutions are
x=3 and x=32

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