$\log_{10}(x^2-x-7)=0.1$
$\log_{10}(x-8)=1-\log_{10}(x+1)$
$\log_{10}(x+9)=1+\log_{10}(x+1)-\log_{10}(x-2)$
Note: I solved them as follows:
$x = 3, -2$ but the textbook i'm using said there was no solution for $x$
$x=-2, 9$ but the same textbook simply said $x=9$
Please, can anyone help me simplify the three equations in detail, and if possible, tell me where i might have made a mistake?
For $(2)$ and $(3)$ it looks like using the logarithm property $\log(a)+\log(b) = \log(ab)$ will help. For example on $(2)$, $$\begin{align}\log_{10}(x-8)=1-\log_{10}(x+1) \\ \implies \log_{10}(x-8)+\log_{10}(x+1)=1\end{align} \\ \implies \log_{10}((x-8)(x+1))=1 \\ \implies (x-8)(x+1)=10$$
Where on $(3)$ you can use the fact that $1 = \log_{10}(10)$ to rewrite the equation as $$\begin{align}\log_{10}(x+9)=1+\log_{10}(x+1)-\log_{10}(x-2) \\ \implies \log_{10}(x+9)+\log_{10}(x-2)=\log_{10}(10)+\log_{10}(x+1) \\ \implies \log_{10}((x+9)(x-2))=\log_{10}(10x+10) \end{align} $$ Can you take it from here? Note that once you have found your solutions to each problem, plug $x$ back in to your equation and make sure everything works okay. By that, I mean if (and I'm making this up) you had found that $x=7$ was one solution to problem $2$, then you would have $\log_{10}((7)-8)=1-\log_{10}((7)+1)$, but this is not okay, as $\log_{10}((7)-8) = \log_{10}(-1)$ and the logarithm is not defined for numbers less than or equal to zero, so you'd have to toss the solution $x=7$.
For 3. you can exponentiate both sides and then solve the equation:
$\log_{10}(x+9)=1+\log_{10}(x+1)-\log_{10}(x-2)$
$\iff 10^{log_{10}(x+9)}=10^{1+\log_{10}(x+1)-\log_{10}(x-2)}$
$\iff x+9=10^1\cdot10^{\log_{10}(x+1)}\cdot 10^{-\log_{10}(x-2)}$
$\iff=x+9=\frac{10\cdot(x+1)}{(x-2)}\iff (x+9)(x-2)=10x+10\iff x^2+7x-18=10x+10$
$\iff x^2-3x-28=0 \iff (x-7)(x+4)=0$
$\implies x=7$ as the only real solution
