+26388 how do i solve: ^2log(-2x+3) - ^2log(x+1)>0
rearrange:
\(\begin{array}{rcll} ^2log(-2x+3) - ^2log(x+1) &>& 0 \\ \log_{2}{ (-2x+3) } -\log_{2}{ (x+1) }&>& 0 \\ \log_{2}{ \left( \frac{-2x+3}{x+1} \right) } &>& 0 \quad & | \quad 2^{()}\\ \frac{-2x+3}{x+1} & > & 2^0 \quad & | \quad 2^0 = 1\\ \frac{-2x+3}{x+1} & > & 1\\ \end{array}\)
\(\boxed{~ \begin{array}{rcll} \dfrac{-2x+3}{x+1} & > & 1 \qquad D = R ~ \backslash ~ \{-1\}\\ \end{array} ~}\)
\(\small{ \begin{array}{ccc|ccc} \hline \text{denominator } x+1 > 0 && & \text{denominator } x+1< 0 && \\ \\ \hline -2x+3~ >~ 1\cdot(x+1) &&& -2x+3 {\color{red}~<~} 1\cdot(x+1)&& \\ \\ \begin{array}{ccc|c} x+1 &>& 0 & -1 \\ -2x+3 &>& x+1 & +2x-1\\ \hline \end{array} &&& \begin{array}{ccc|c} x+1 &<& 0 & -1 \\ -2x+3 &<& x+1 & +2x-1\\ \hline \end{array} \\\\ \begin{array}{ccc|c} x &>& -1 & \\ 2 &>& 3x & :3\\ \hline \end{array} &&& \begin{array}{ccc|c} x &<& -1 & \\ 2 &<& 3x & :3\\ \hline \end{array} \\\\ \begin{array}{ccc|c} x &>& -1 & \\ \frac23 &>& x & \\ \hline \end{array} &&& \begin{array}{ccc|c} x &<& -1 & \\ \frac23 &<& 3x & \\ \hline \end{array} \\\\ \mathbf{ -1 < x < \frac23 }&& & \text{no solution}&& \end{array} }\)
hello, there were still things i couldnt understand.
for example the first part of the answer were u had put in that D=R \{-1}...I would like to know how you got -1 as answer.
Further on i would like to know were f(x)-g(x)>0 can be found in a graphic (if this exist and is possible)
+26388 Hallo Guest
The denominator must not 0 (null).
The denominator is 0, if
\(\begin{array}{rcll} x+1 &=& 0 \quad & | \quad -1\\ x+1-1 &=& 0-1 \\ x+0 &=& -1 \\ x &=& -1 \end{array}\)
so \(x=-1\) is not permissible or \(D = R ~ \backslash ~ \{-1\}\)
\(\dfrac{-2x+3}{x+1} > 1\)
+26388 Hallo Guest
The denominator must not 0 (null).
The denominator is 0, if
\(\begin{array}{rcll} x+1 &=& 0 \quad & | \quad -1\\ x+1-1 &=& 0-1 \\ x+0 &=& -1 \\ x &=& -1 \end{array}\)
so \(x=-1\) is not permissible or \(D = R ~ \backslash ~ \{-1\}\)
\(\dfrac{-2x+3}{x+1} > 1\)
Hello,
i see that u had put 2 solutions for the denominator in 2 different ways, but on the right side and the last part....does 3x still remain unsolved after u had divided it by 3
2 < 3x | :3
2/3 < 3x
does the (x) still remain there? or should it look like this:
2 < 3x | :3
2/3 < 3
