MHB Finding $a$ and $b$ for $f(x)=|\lg (x+1)|$

[anemone]

Suppose $f(x)=|\lg (x+1)|$ and real numbers $a$ and $b$ where $b>a$ satisfy $f(a)=f\left(-\dfrac{b+1}{b+2}\right)$ and $f(10a+6b+21)=4\lg 2$.

Find the values of $a$ and $b$.

 

[solakis1]

Suppose $f(x)=|\lg (x+1)|$ and real numbers $a$ and $b$ where $b>a$ satisfy $f(a)=f\left(-\dfrac{b+1}{b+2}\right)$ and $f(10a+6b+21)=4\lg 2$.

Find the values of $a$ and $b$.

[sp]is the answer a=0 b=-1[/sp]

 

[anemone]

Sorry, solakis, those are not the values for $a$ and $b$.

 

[Opalg]

Spoiler

If $f(x)=f(y)$ then $\log(x+1) =\pm \log(y+1)$. Apart from the obvious solution $y=x$, that has the solution $y+1 = \dfrac1{x+1}$. Given that $f(a) = f\left(-\dfrac{b+1}{b+2}\right)$, it follows that either $-\dfrac{b+1}{b+2}=a$ or $-\dfrac{b+1}{b+2}+1 = \dfrac1{a+1}$. The second of those two possibilities leads to solakis's suggestion $a=0$, $b=-1$, which would be a solution apart from the fact that it does not satisfy the condition $b>a$.

So we have to look at the possibility $a = - \dfrac{b+1}{b+2}$. The equation $f(10a+6b+21) = 4\log2$ then says that $\log\left(-\tfrac{10(b+1)}{b+2} + 6b + 22\right) = \pm\log16$. Therefore $-\tfrac{10(b+1)}{b+2} + 6b + 22 = 16 $ or $\tfrac1{16}$.

If $-\tfrac{10(b+1)}{b+2} + 6b + 22 = 16 $ then $(b+1)\bigl(-10+ 6(b+2)\bigl) = 0$. That again has the unwanted solution $b=-1$, but it also has the solution $b+2 = \frac{10}6$, from which $b=-\frac13$. The corresponding value of $a$ is $-\frac25$.

(There was also the possibility that possibility that the equation $-\tfrac{10(b+1)}{b+2} + 6b + 22 = \tfrac1{16} $ might lead to a solution. But in fact that equation has no real solutions.) So the solution is $\boxed{a=-\dfrac25,\ b=-\dfrac13}$.