Mathematics - Functions Question with Solution | TestHub

First note that $2x+3>0$ and $2x+3\ne 1$, that is, $x>-3/2$ and $x\ne -1$. Also, $3x+7>0$ and $3x+7\ne 1$, that is, $x>-7/3$ and $x\ne -2$. Suppose $x>-3/2,x\ne -1$. Then the given equation can be written as:

$$\frac{\log [(2x+3)(3x+7)]}{\log (2x+3)}=4-\frac{2\log (2x+3)}{\log (3x+7)}$$

$$1+\frac{\log (3x+7)}{\log (2x+3)}=4-\frac{2\log (2x+3)}{\log (3x+7)}$$

Put $y=\frac{\log (3x+7)}{\log (2x+3)}$. Then $1+y=4-\frac{2}{y}$.

Therefore:

$$y=3-\frac{2}{y}$$

$$y^2-3y+2=0$$

$$(y-1)(y-2)=0$$

This gives $y=1$ or $y=2$.

Case 1: Suppose that $y=1$. Then $\frac{\log (3x+7)}{\log (2x+3)}=1$, which implies $\log (3x+7)=\log (2x+3)$.

So, $3x+7=2x+3$, which gives $x=-4$. This is rejected because $x>-3/2$.

Case 2: Suppose that $y=2$. Then $\frac{\log (3x+7)}{\log (2x+3)}=2$.

So, $\log (3x+7)=2\log (2x+3)=\log (2x+3{)}^2$.

Therefore, $3x+7=(2x+3{)}^2$.

$3x+7=4x^2+12x+9$.

$4x^2+9x+2=0$.

$(4x+1)(x+2)=0$.

This gives $x=-1/4$ or $x=-2$. Here $x=-1/4$ (since $x>-3/2$).

so -4x = 1